jmc

There are 8 choices for seats for Alice to sit in. Once Alice is seated, there are 5 seats left for Bob, since he won't sit in either seat immediately next to Alice. This leaves 6 people to place in the remaining 6 seats, which can be done in $6!$ ways. However, we must divide by 8 to account for the 8 rotations of the table. So the number of arrangements is $\frac{8\times 5\times 6!}{8}=5\times 6!=\boxed{3600}$. Alternatively, we can account for the rotations at the beginning, by fixing the table around Alice. Bob can't sit in her seat, or the two seats next to her. This leaves 5 places for him to sit. Then, this leaves 6 unique seats for the 6 remaining people, so there are 6! ways to sit them after Bob is sitting. So the answer is $5\times 6!=\boxed{3600}$.