In this problem, you have to add an alkyl halide, specifically HBr, to an alkene. As we all know, when an alkylhalide adds to an alkene, a carbocation intermediate forms which means you have to look out for possible rearrangements such as a methyl shift or a hydride shift. The reaction is regioselective.
The top problem, however, is not regioselective; it will give both products, not one is formed over the other because both have the same stability.
In the second problem, the reaction is regioselective and we see a rearrangement. Without a rearrangement, the Br would have added to carbon 2, a secondary carbocation. A rearrangement is possible, a hydride shift. By moving a hydrogen atom from carbon 3 to carbon 2, you will get a tertiary carbocation on carbon 3, which is more stable than a secondary carbocation. As previously stated, the reaction is regioselective and wants to be as stable as possible. The hydride shift and addition of Br to carbon 3 and H to carbon 1 gives the most stable product, the major product, which is shown.
The third problem is similar to the second one, however instead of a hydride shift, a methyl shift is seen. A methyl group moves from carbon 3 to carbon 2, just as the hydride did in the previous problem. Once again, this occurs to give the most stable carbocation for the halide to add to and, in turn, the most stable product, or the major product.
The final problem shows a hydride shift as well. This is to get the most stable carbocation for the halide to add to. So once H adds to the less substituted carbon of the double bond, a hydride ion moves from carbon 3 to carbon 2 to form a tertiary carbocation on carbon 3 for the Br to add on to.