Here is an post-mortem analysis / "how to" for the MT. The questions are split by the sections. At the start of each section are a few suggestions of what to look for or how to tackle the question type.

RELATIVE PROPERTIES:
Identify the controlling feature, which is not always as obvious as it may appear. Look for two pairs of similar systems to compare that have minimal differences in structure. If a compound is named, draw it out. If a reaction is involved, identify the type of reaction and then what the controlling factors are.

Qu1: B
Reaction is about additions of HX to alkenes or alkynes. These reactions give the Markovnikov product via the most stable carbocation intermediate. The rate of the reaction is controlled by the formation of the carbocation and the variable here is the pi system and hence the carbocation intermediate. i is an alkene and gives a tertiary carbocation, ii is an alkyne while iii is an alkene and gives a secondary carbocation. Alkynes are less reactive since they "would require" the formation of the less stable vinyl carbocations and hence are forced to go via a less favourable termolecular process. So i > iii > ii.

Qu2: E
Vinylic hydrogens are found on the sp2 C within alkene C=C units. So, i has 1, ii has 0 and iii has 2. Hence iii > i > ii.

Qu3: A
Carbocation stability.... due to (a) alkyl groups, which are weak electrons donors, and (b) resonance with O lone pairs. These effects add stability due to charge delocalisation.  i is secondary with an O attached that can stabilise the charge via resonance using the O lone pair, ii is a simple secondary carbocation and iii is secondary with no resonance stabilisation (the sp3 center prevents the resonance)....the electronegative effect of O through induction will destabilise it. Hence i > ii > iii.

Qu4: A
Reaction is the Diels-Alder reaction and we are looking at the dienes reacting with the dienophile methyl acrylate (drawn in the question).  The reactivity increases in the normal Diels-Alder reaction with electron donating groups on the diene.  i has an alkyl group (a weak electron donor) on the diene C=C, ii has an alkyl group but it is not on the C=C so it has no effect, and iii can not achieve the reactive s-cis conformation because the ring locks it in the unreactive s-trans conformation. Therefore the reactivity is i > ii > iii.

Qu5: AB
The reaction is the radical addition of HBr to alkenes. First draw out the starting material, 1-methylcyclopentene. iii and ii have correctly added the Br and H across the C=C therefore iii has the highest yield since it is the anti-Markovnikov product, while ii is the Markovnikov product. Since the radical addition of HBr to alkenes gives the anti-Markovnikov product as the major product then we get iii> ii > i.

Qu6: D
An application of acidity. The functional groups are i is a carboxylic acid, ii a terminal alkyne, and iii an alcohol. Carboxylic acids are the most acidic since the carboxylate ion puts the negative charge on an electronegative O where there is resonance delocalisation. In an alcohol, the negative charge is localised on the O atom. In an alkyne, the negative charge is localised on the sp C atom (remembering what bases are used to deprotonate alkynes is also useful here - alkoxides don't work). Or we could use the pKas... alcohol =15, alkyne = 25 and carboxylic acid = 5: the lower the pKa the stronger the acid. So we have in terms of pKa ii > iii > i.

Qu7: D
The reaction is the addition of hydrogen (H2) to alkenes or alkynes. These reactions are controlled the the strength of the pi bond. So alkynes are more reactive than alkenes because the pi bond in the triple bond is weaker. Since the trans alkene is more stable than the cis, the cis isomer is more reactive. So that means in terms of reactivity, ii > iii > i.

Qu8: B
The reaction is the dihydroxylation of an alkenes. It's a syn addition. First step is draw out the starting material, (Z)-pent-2-ene (i.e. cis) i and iii have correctly added the two -OH groups across the C=C therefore i has the highest yield since it is the syn addition product while iii is the anti product. So we get i> iii > ii.

Qu9: B
The reaction is the E2 elimination of an alkyl halide. For cyclic systems this means that the LG (Br needs to be axial). The large t-butyl group has a very strong preference to be in the more stable equatorial position. This means that in i when the t-Bu is equatorial, the Br is already axial and readily eliminates. In ii when the t-Bu is equatorial, so is the Br and because the ring flip is difficult, the elimination is very slow. In iii the ring flip to put the Br in the required axial position is feasible and the elimination can occur. So, in terms of rate of reaction, i > iii > ii.

Qu10: A
Try the C=C in different locations, pay attention to valence rules and symmetry. i has 6, ii has 4 and iii has just 2, so i  > ii > iii 

STARTING MATERIALS AND PRODUCTS:
If you are trying to find the product, then you should probably just work forwards through the sequence of reactions.
If you are looking for the starting material, then working backwards is probably the best way to go....
Basically depends on the need to know and identify the reactions, this is often triggered by looking at the functional groups in the molecules.

Qu11: E
Working forwards... the alkene reacts with the aqueous bromine to give the 1,2-bromohydrin which then reacts with the base to give an epoxide. The epoxide then undergoes an acidic ring opening with methanol as the nucleophile (attacking the more substituted carbon) to give the trans 1,2-system (look at the -OH / -OCH3). A has the wrong regiochemistry (it's 1,1-), B has the wrong substituents and regiochemistry, C has the wrong substituents and stereochemistry (it's cis), and D has the wrong stereochemistry.

Qu12: B
Working forwards..... this is hydration of the alkyne to give a methyl ketone alcohol followed by an E2 elimination of an alkyl halide to give the conjugated ketone. This then acts as the dienophile in a normal Diels-Alder reaction with 1,3-cyclopentadiene. A has the wrong dienophile, C has the wrong diene and dienophile, D and E have the wrong dienophile (they would require an alkyne).

Qu13: C
The alkyne needs to be alkylated twice via SN2 with the nucleophilic terminal acetylide by using the NaNH2 then the alkyl halides (once with an ethyl halide and once with a methyl halide) to make the unsymmetrical alkyne which then under goes a dissolving metal reduction to give the trans-alkene. A and D have the wrong reduction (gives the cis-alkene), B gives a symmetrical alkene (uses the wrong alkylating agent), E doesn't work because you have to alkylate the terminal alkyne before you reduce to the alkene.

Qu14: B
Working forwards..... the epoxide undergoes an acidic ring opening with water as the nucleophile (attacking the more substituted carbon) to give the trans 1,2-system, followed by dehydration to give the conjugated diene. Addition of HBr under kinetic control (Markovnikov, via more stable carbocation) gives the allylic bromide where the C=C then undergoes the Simmonds-Smith reaction to give the cyclopropane.

Qu15: E
In order to introduce an extra double bond, we need to install a LG and then eliminate. The key here is to think about the regiochemistry of the bromination steps and then the elimination reactions.

Qu16: A
Working backwards...reactions look like alkene ozonolysis with a reductive work-up and therefore reconnect the aldehydes to reveal the substituted cyclohexene that was formed via the Diels-Alder reaction in step 1. Therefore we are looking for the dienophile which needs to be a cis-dicarboxylic acid.

Qu17: E
Working backwards...the product is an ether, which can be made by an alcohol reacting as a nucleophile in a substitution reaction where there is a leaving group. Here the alcohol has been made via the hydroboration-oxidation of an alkene to give the anti-Markovnikov product (at the end of the terminal end of the alkene). The leaving group is a tosylate (step 1). Since tosylates are made from alcohols, only D and E are viable starting materials. The alkene in E is required for the hydroboration step.

Qu18: C
Working backwards...the product is an ether, which looks to have been made via an alkoxy-mercuration of an alkene (i.e. an alcohol reacting with the more substituted end of a cyclic mercurinium ion formed by reaction with an alkene). The alkene has been formed by an E2 elimination of an alkyl halide (the base suggests this) where the alkene includes the C where the alkyl chain attaches to the ring.

REGIO- and STEREOCHEMISTRY:
How well do you know your reagents ? Look at what has actually happened in terms of the reaction functional group transformation and then first look for any regiochemical issues then finally the stereochemistry last (it's the hardest to sort out).  In cases where more than one product is formed in equal amounts (e.g. the enantiomers), then both must be selected for full marks, part marks are given when only one of the pair is selected.

Advice : in each case draw the starting material in the conformation in which is reacts or the product in the conformation in which it is initially formed using wedge-hash diagrams. It is a good idea to draw the materials in such a way that the new bonds are in the plane of the page.  Once you have drawn the materials it may also be good for you to use model kits for these questions too.  Once you have drawn the materials in this way, you may need to consider rotations around sigma bonds to make your answer match the options.  An alternative approach could be to assign configurations to your drawn answer to compare them with the options - this can be slow and prone to error.

Qu19: B
Working forwards...the starting material is an alkyne, the first set of reagents indicate the catalytic hydrogenation reduction of alkyne to an alkene which is a syn addition giving the cis-alkene. This is then followed by epoxidation and with the same stereochemistry as the original alkene. This is followed by acidic ring opening using aq. acid. This means that the reaction is SN1 like and the water attacks the more substituted C in the epoxide.... hence we end up with the overall anti addition to give the 1,2-dihydroxy. Need to convert this to a Fischer projectionnote that the arrangement of the alkyl chain in the cis alkene is similar to that of the Fischer projections. A and E have the wrong stereochemistry. C and E have the wrong regiochemistry.

Qu20: D
Working forwards...the starting material is an alkene, which reacts with the aqueous bromine to give the 1,2-bromohydrin via an anti addition, where the -OH ends up at the more substituted (tertiary) position due to the +ve character of the intermediate halonium ion. A has the wrong stereochemistry. B, C and E have the wrong regiochemistry.

Qu21: A
Working forwards, the strong base eliminates the bromide via E2 elimination of the alkyl halide giving the conjugated diene which then undergoes 1,4-addition of chlorine (thermodynamic control). B, C, D and E have the wrong regiochemistry. B is the kinetic product.

Qu22: C
Working forwards, we have an intramolecular Diels-Alder reaction. Key parts to notice would be the diene "S" is part of a five membered system, the dienophile is trans and the location of the methyl group on the diene relative to the other alkyl chain that links the diene and the dienophile.

Qu23: A
Working forwards, we are brominating an alkene to a 1,2-dibromide, followed by elimination (x2) to get an alkyne which then undergoes hydroboration-oxidation of an alkyne (anti-Markovnikov) to give the aldehyde.

Qu24: A
Working forwards... we have addition of HCl to an alkyne, followed by HBr to the alkene.... this will give the geminal (1,1-) dihalide,with the halogens at the benzylic position (due to cation stability).

Qu25: C
Working backwards... we have ozonolysis of an alkene with an oxidative work up.... the four ketones come from tetrasubstituted alkenes. A, B, D and E would give carboxylic acids.

PI SYSTEMS:
All about knowing properties of pi systems such as resonance, structure, bonding, shapes, and terminologies etc.

Qu26: CDE
Conjugation requires an extended pi system (at least 3 atoms) where the pi systems can interact (parallel, not perpendicular). This is not restricted to all carbon systems. Make sure you draw out all the bonds to be sure. A is an isolated diene. B is an alkene. C is a conjugated diene. D has an alkene conjugated with a carbonyl group cumulated diene. E is an allylic anion.

Qu27: E
Resonance contributors are derived via the delocalisation of the pi electrons with the pi system and can be derived by pushing curly arrows. Use the rules for recognising resonance structures. A, B and D can be ruled out because the +ve charge can't be located on a C that was sp3 in the original structure.

Qu28: C
The key facts here are (1) conjugated dienes are more stable than isolated dienes (due to conjugation stabilisation), (2) in acyclic systems, trans alkenes are more stable than cis (sterics) and (3) in conjugated dienes the s-trans is more stable than the s-cis (sterics). Therefore the least stable system is C where we have an isolated diene with a cis double bond. A, B and D are conjugated dienes. E is an isolated diene but has a trans double bond..

Qu29: C
The least exothermic heat of hydrogenation means we are looking for the most stable isomer. The key facts here are (1) alkyl groups stabilise alkenes so more substituted alkenes are the more stable, (2) conjugated dienes are more stable that isolated dienes which in turn are more stable than alkynes and (3) in conjugated dienes the s-trans is more stable than the s-cis (sterics). Therefore the most stable system is C where we have a conjugated alkene with a trans alkene. A and B are alkynes. E is an isolated diene. D is a cis-alkene.

Qu30: B
Resonance contributors are derived via the delocalisation of the pi electrons with the pi system and can be derived by pushing curly arrows. Use the rules for recognising resonance structures. A, C and D can be ruled out because the H atom has moved from C to O. E is wrong because it has a different overall charge.

Qu31: C
The reaction is a dissolving metal reduction to give the trans-alkene. This only works for alkynes...

Qu32: D
This is the addition of HCl to alkenes or alkynes. These reactions give the Markovnikov product via the most stable carbocation intermediate....the reaction that is the fastest will be the one that proceeds via the most stable carbocation. A and E give secondary carbocations, B gives a tertiary carbocation, C : alkynes are less reactive since they "would require" the formation of the less stable vinyl carbocations and hence are forced to go via a less favourable termolecular process. D would give a tertiary benzylic carbocation.

Qu33: A
We need to look for the s-cis conformation at C2-C3 of the 3E (trans) isomer....B is 3Z, C is not a pentadiene, the longest chain is C4, D is s-trans at C2-C3 and E is not a 2-methyl system.

Qu34: C
Resonance contributors are derived via the delocalisation of the pi electrons with the pi system and can be derived by pushing curly arrows. Use the rules for recognising resonance structures. A can be ruled out because an H atom has moved to create the terminal C=C, B, D and E are wrong because the radical is now at a C that was sp3, so an H would have had to move.