Firstly, label them A,B,C,D,E. The question is intentionally pissing you off. I assume it means the seats at the table are not unique, that is A,B,C,D and E is the same as B,C,D,E and A.

a) This is a simple 4!, since position doesn't matter.

b) By this, I assume it means directly right. As such, there should be 3! ways to do it.

c) If the 3 people are defined, there are only 2 other people to be selected. but, brad and bob can be on either side, so there is actually 2*2 = 4.

d) With this, you can use b). If there are 3! for right, there are 3! for left. This is how many ways someone can sit next to someone. Take this away from 4!

d) Consider here the gaps. Between A and B, there must be one 1 person gap and 1 2 person gap. The 2 person gap must be C and D, which have 2 ways to be positioned. The other gap must only have E. As such, there are 2*2

Take with a pinch of salt though - I may have just been entirely wrong.