Your answer to the first question is correct. Here is another way to see it: In a circle, if the three women sit together, so do the four men. Thus, there are two blocks, which can be arranged in $(2 - 1)! = 1! = 1$ way around the table. Within the block of women, the women can be arranged in $3!$ ways. Within the block of men, the men can be arranged in $4!$ ways. Hence, there are $3!4!$ arrangements of three women and four men around a circular table in which all the women sit together.
For the second problem, seat the men first. Since the four men are sitting in a circle, there are $(4 - 1)! = 3!$ distintinguishable arrangements of the men. This leaves four spaces in which to place the three women, one to the right of each man. We can select three of these four spaces in $C(4, 3)$ ways, and arrange the women within the selected spaces in $3!$ ways. Hence, the number of possible seating arrangements of four men and three women around a circular table in which no two of the three women sit together is $3! \cdot C(4, 3) \cdot 3! = 6 \cdot 4 \cdot 6 = 144$.
Your error in the second problem was to apply the rule $(n - 1)!$ twice. Once the men are seated, rotating the women produces a new arrangement. To see this, suppose Andrew, Bruce, Charles, and David are seated around the table in counterclockwise order. Suppose Elizabeth sits to the right of Andrew, Fiona sits to the right of Bruce, and Gretchen sits to the right of Charles. If each women moves to her right (past a man), Elizabeth is now to the right of Bruce, Fiona is to the right of Charles, and Gretchen is now to the right of David. Clearly, this is a different arrangement.
I would do the problem exactly the way you did it.
Here is an alternative approach to confirm your answer.
We can line up the five students in $5!$ ways, leaving spaces between them and at the ends of the row in which to insert the teachers. There are six such spaces, four between successive students and two at the ends of the row. We can insert the three teachers in $P(6, 3) = 6 \cdot 5 \cdot 4$ ways. This gives us $$5! \cdot 6 \cdot 5 \cdot 4$$ linear arrangements of students and teachers in which no two teachers are consecutive.
However, since we wish to arrange the students and teachers around a circular table so that no two teachers sit in consecutive seats, we must exclude those linear arrangements in which teachers are at both ends of the row. There are three ways to select the teacher at the left end of the row, two ways to select the teacher at the right end of the row, and four ways to place the remaining teacher in one of the four spaces between successive students. Hence, there are $$3 \cdot 2 \cdot 4 \cdot 5!$$ linear arrangements in which teachers are at both ends of the row.
Hence, there are $$6 \cdot 5 \cdot 4 \cdot 5! - 3 \cdot 2 \cdot 4 \cdot 5! = (120 - 24)5! = 96 \cdot 5!$$ linear arrangements of teachers and students so that no two teachers are consecutive and teachers are not at both ends of the row.
These linear arrangements correspond to the permissible ways we can seat the students and teachers around the table. To account for rotational invariance, we divide the number of linear arrangements by $8$, which yields
$$\frac{96 \cdot 5!}{8} = 12 \cdot 5! = 1440$$
permissible seating arrangements around a circular table, as you found.
Best Answer
You can seat Mrs Jones in any of $8$ places. Once she’s seated, there are $3$ places that are not available to Mr Jones: the place where she’s seated, and the two adjacent places. Thus, there are $5$ choices for his seat. That leaves $6$ people who can fill the remaining $6$ seats in any of $6!$ possible orders, for a total of $8\cdot5\cdot6!=28,800$ possible arrangements if the seats are the table are individually identified.
If we care only about who sits next to whom, then we don’t care which of the $8$ seats Mrs Jones takes: all we care about is where the others sit in relation to her. That reduces the number of choices to $5\cdot6!=3600$ and must therefore be the intended interpretation.