For #1: Choose whether A is to the left of B or to the right of B (2 possibilities). Then, choose the position of the person on the left (A*B****, *A*B***,**A*B**,***A*B*,****A*B) and the five choices for when B is on the left. So, there are 10 possible ways for A and B to be positioned. Regardless of how A and B are positioned, there are 5 seats left. There are 5! ways to seat the remaining people. So, there are ways of seating all seven people.
For #2: Don't count permutations where the 1 and 2 are not together. What you said about combining the 1 and 2 is correct. First, choose the order of the 1 and 2. Like you said, there are 2 possibilities. Then, you are permuting 8 "things". So, the answer is .
For #3: b) You have 3 people. In the left-most chair, put one of them. In the right-most chair, put one of them. Then, choose of the remaining eight chairs to put the last person. So, that is .
c) If the two end chairs are empty, then you only have 8 chairs. So, that is ways to seat three people so that the end chairs are empty.
For #4: b) Count how many are larger than 400. That means the first digit must be 5, 7, or 9. So, the first digit has 3 choices. The second and third digits each have 5 choices. That is 3-digit numbers larger than 400. Next, count how many are larger than 350 but less than 400. That means the first digit must be 3, the second digit may be 5, 7, or 9, and the last may be anything. That is 3-digit numbers. All together, there are 3-digit numbers larger than 350 using only odd digits. Note that this is not a permutation problem. Digits may be repeated.
Edit: For #4 a) Digits may be repeated. So, there are 3-digit numbers using only odd digits.