8 people sit around a round table. Arrangements are considered the same if the same student is to the right/left. So, in total there are $7! = 5040$ ways to seat.
Part 1: How many arrangements are there with B sitting next to both A and C?
- For this one, I think that B would have to be in the middle, and there are 2 ways to place B between A and C. So, $2!\cdot5! =$ 240 ways.
Part 2: How many arrangements are there with B sitting next to A or C?
- For this one, I was thinking about subtracting from the opposite: B will sit next to anyone other than A/C.
- So, $7! - 6 \cdot 1\cdot 4\cdot 5!=$ 2160 ways.
Can anyone tell my approach looks good / verify my answers?
Your solution to part $1$ is correct.
When counting the arrangements where $B$ is next to neither $A$ nor $C$, there are...
$5$ choices for the person sitting on $B$'s left; anyone but $A$, $B$ or $C$. [You had $6$ for this number, which was your mistake.]
$4$ choices for the person sitting on $B$'s right; anyone but $A$, $B$, $C$, or the person sitting on $B$'s right.
$5!$ ways to choose everyone else; $5$ choices for the person $2$ places to the right of $B$, $4$ choices for the person $3$ places to the right, etc.
Finally, the answer is $\boxed{7!-5\cdot 4\cdot 5!}$.