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:
Alkenes and alkynes undergo electrophilic addition with acids which are controlled by carbocation stability and the nature of groups attached to the pi unit. Alkynes are less reactive since they would require the much less favourable termolecular pathway to avoid the formation of the very unfavourable vinyl carbocations. The ease of carbocation formation dictates the rate of reaction. In the case of alkynes, the vinyl C+ would be too unstable and a slower termolecular mechanism occurs, so the alkyne is the slowest. The more highly substituted alkene will give the more highly substituted C+ (tertiary) and hence reacts the fastest.

Qu2:
Use the pKas of the acids: terminal alkyne CH = 25, ammonia NH = 35 and alcohol = 15 or think about the bases that cab be used to prepare an acetylide ion !

Qu3:
Reaction is the Diels-Alder reaction and we are looking at the dienophiles reacting with the diene (1,3-cyclopentadiene).  The reactivity increases in the Diels-Alder reaction with electron withdrawing groups on the dienophile. The carbonyl group (such as in an ester) is electron withdrawing through resonance (and two are better than one) while alkoxy groups (as in ethers) are electron donating through resonance.

Qu4:
The acid catalysed hydration of an alkene adds the HO and H across the C=C in a Markovnikov fashion. Here we are asked about 3,3-dimethylbut-1-ene. The major product is the rearranged product since the reaction does proceeds via an carbocation, followed by the simple Markovnikov product and then the anti-Markovnikov product.

Qu5:
Alkenes undergo electrophilic addition. In the reactions with HX, the rate is controlled by the rate of carbocation formation. In this case, this means the acidity of the HX species and hence the stronger the acid the faster the carbocation is created.

HBr > HCl > H₂O

Qu6:
This question relates directly to the sodium borohydride reactivity experiment from the laboratory where the reactivity aldehydes > ketones > esters.

Qu7:
The reaction is alkyne hydration. and its regioselectivity. Pentan-2-one is a methyl ketone. Aq. hydration of a simple terminal alkyne typically gives methyl ketones as the major product ("Markovnikov"). while simple unsymmetrical internal alkynes give mixtures of products. In contrast, hydroboration / oxidation of simple terminal alkyne typically gives aldehydes as the major product ("anti-Markovnikov").

Qu8:
Carbocation stability.... Look at the C atom bearing the charge and what's attached to it. The vinyl carbocation is the least stable and the allylic carbocations is the most stable since it is resonance stabilised by a C=C. Note that the Br can also stablise the +ve charge but the size of the Br atom and the length of the C-Br bond mean the required orbital overlap is not as effective.

Qu9:
To count vinylic H we need to count the H attached to the C atoms that are part of the double bonds.

Qu10:
All the bonds are C-C bonds. The factors involved are the hybridisation of the C atoms involved. C-C bonds involving two sp2 C atoms are shorter than two sp3 C atoms (due to higher s character).

STARTING MATERIALS, REAGENTS 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:
Working forwards... the product contains an alkyne and we a starting from an alkene. Need to have 2 leaving groups to make 2 pi bonds, so alkene bromination followed by strong base to eliminate.

Qu12:
Working forwards...catalytic hydrogenation of the alkene to the alkyl group followed by elimination of the alkyl bromide to give but-1-ene. This is followed by hydroboration/oxidation which gives the anti-Markovnikov alcohol.

Qu13:
Working backwards...The final product is a trans-alkene which has been formed by a dissolving metal reduction of an alkyne. Steps 1 and 2 look to be alkylation of a terminal alkyne.

Qu14:
Working forwards....starting from the terminal alkyne we need to lengthen the C chain by adding 4 more C atoms (i.e. alkyne alkylation) and then convert the terminal alkyne into a methyl ketone.

Qu15:
Working fowards..... the starting material is C4 and the product a C3 carboxylic acid. We could make the carboxylic acid while reducing the chain length by using ozonolysis of an alkene or alkyne. So elimination then ozonolysis.

Qu16:
Working forwards...radical addition of HBr to the alkene will give the anti-Markovnikov bromide. Treatment with a weak base will cause an intramolecular nucleophilic substitution to give the cyclic ether. Tip... count and number the C atoms to get the right ring size and positioning of the groups.

Qu17:
Working forwards....starting from the terminal alkyne we need to lengthen the C chain by adding 4 more C atoms (i.e. alkyne alkylation) and then convert the terminal alkyne into an aldehyde.

Qu18:
Working forwards... oxymercuration / demercuration of the alkene will give the Markovnikov alcohol but avoid the problem of rearrangement.

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:
Addition of HBr to the diene at elevated temperature will result in the thermodynamic product(s). Note that these products are of essentially equal stability as both of disubstituted trans-alkene and 2^o allylic bromides.

Qu20:
The catalytic hydrogenation is a syn addition, giving the cis-alkene. The cis-alkene is reacted with basic osmium tetroxide to give a 1,2-diols via a syn addition.

Qu21:
Working backwards, the chloride has been made from the alcohol which in turn has been created by the hydroboration/oxidation (anti-Markovnikov alcohol, B atom adds to the least hindered end) from the alkene. The best reaction is the one that gives the best regioselectivity so the biggest difference in the ends of the alkene.

Qu22:
Halohydration via an anti-addition and putting the HO- at the tertiary position. Then treatment with base forms the epoxide. Reaction of phenyl magnesium bromide, a strong nucleophile, opens the epoxide via an SN2 like reaction, attacking at the least hindered end of the epoxide to give the trans-substituted alcohol.

Qu23:
The product looks to be from a Diels-Alder reaction of 1,3-cyclopentadiene with trans-but-2-ene. The trans-alkene is made by the dissolving metal reduction of the alkyne. There is no way to get this stereochemistry after the Diels-Alder reaction, it is best obtained by using the right dienophile.

Qu24:
The best way to obtained the dibromide with the required stereochemistry is to perform the bromination (an anti addition) on the trans-alkene which would be formed via dissolving metal reduction of an alkyne.

Qu25:
Halohydration via an anti-addition (so -Cl and -OH are trans and on opposite faces of the ring) and putting the HO- group at the tertiary position.

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

Qu26:
Conjugation requires an extended pi system (at least 3 atoms) where the pi systems can interact (parallel orbitals, not perpendicular). The image below has highlighted the p orbitals (blue circles for top down view of vertical p orbitals, and black for those that are perpendicular to the others).

Qu27:
Resonance contributors are derived via the delocalisation of the pi electrons across the pi system and can be derived by pushing curly arrows. Use the rules for recognising resonance structures.

Qu28:
Hydration of alkenes with aq. sulfuric acid is controlled by the rate of carbocation formation where the most stable carbocation forms fastest. Here, the most stable cation that can form is tertiary and allylic.

Qu29:
The key issues are (1) conjugated systems are more stable than isolated systems, (2) alkyl groups on pi bonds stabilise them, (3) trans isomer are more stable than cis isomers (steric effects) and (4) s-trans is typically more stable than s-cis. Therefore the most stable isomer is the conjugated diene that includes two di trans-substituted alkene units.

Qu30:

Qu31:
Tautomers (remember alkyne hydration ?) of ketones are the enols.

Qu32:
The addition of HCl to an alkyne is a concerted termolecular process with H+ adding from one side while Cl- reacts on the other.

Qu33:
Organic cis / trans nomenclature is based on the stereochemistry of the parent chain that forms the root of the name (longest chain that includes the principal functional group).

Qu34:
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. Note that the -ve charge can only be relocated to another atom in the original pi system (i.e. it can't be moved to an atom that was originally sp3).