SN1 mechanism

This is an S_N1 reaction, but the carbocation is special because it's allylic.

If the Br^- nucleophile attacks the 3º resonance form, the 1,2 product is formed. (This is the product that would have formed if resonance was not involved.)

If the Br^- nucleophile attacks the 2º resonance form, the 1,4 product is formed.

This is an S_N1 reaction. We know this because none of the reagents have charges (H₂O is neutral; if it were HO^-, it would probably be S_N2 or E2). So a carbocation will be formed in this reaction, and the compound whose carbocation is the most stable will react the fastest. So the 3º bromide will react fastest with water.

This is an S_N1 reaction, but with a twist.

The top reaction is a regular S_N1 reaction; the leaving group (Br^-) leaves and the nucleophile (CH₃OH) attacks the carbocation.

The twist is that the carbocation is allylic, and has resonance, so there is another carbon that the nucleophile can attack. 

So instead of just one S_N1 product, two products are formed. The product that doesn't involve resonance is called the direct addition or 1,2 product. The product that involves allylic resonance is called the conjugate addition or 1,4 product.

This is a substitution reaction. Because the reaction takes place in acid and the leaving group is on a 2º carbon, it will probably form a carbocation (S_N1 mechanism).

^-OH is a poor leaving group so the alcohol will protonate first, so it can leave as H₂O, which is neutral.

The 2° carbocation formed immediately undergoes a 1,2-methyl shift (a rearrangement) to form the more stable 3° carbocation, so the product is the 3° alkyl chloride instead of the 2º alkyl chloride, which would have formed in the absence of rearrangement.

For a), the product is neutral and so you are done after the nucleophile (Cl-) attacks the carbocation.

For b), the the intermediate is a protonated alcohol, and so you must do a proton exchange step (also called a hydrogen exchange or deprotonation) to get the final alcohol product, which is neutral.

For c), there are two different types of beta hydrogens, and so two different alkenes are possible.

The more stable alkene is the one that will form, and this will always be the most highly substituted alkene. This is Zaitsev's rule. The rationale for this is hyperconjugation: neighboring carbon atoms stabilize an alkene.