If you want to find the number of ways to arrange men and women so that we have no men or women sitting consecutively we get
$$2\cdot5!\cdot5!$$
Since There $5!$ ways to arrange the males skipping a seat in between and $5!$ ways to arrange the females in the remaining $5$ spots. Then there are $2$ ways to start the row, either male or female.
Also if you're trying to find the number of ways to only group men together and women together (MMMMMWWWWW and WWWWWMMMMM) the number is the same, $2\cdot5!\cdot5!$
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.
Best Answer
Two circular seating arrangements are considered to be distinguishable if the relative order of the people differs.
Method 1: Consider the diagram below.
We can describe a particular seating arrangement by describing the order of the men as we proceed counterclockwise (anticlockwise) around the circle. Notice, however, that we can describe this particular seating arrangement in six ways, depending on whether we start with A, B, C, D, E, or F. Thus, this particular circular arrangement corresponds the six linear arrangements ABCDEF, BCDEFA, CDEFAB, DEFABC, EFABCD, FEABCD. More generally, each circular arrangement of the six men corresponds to six linear arrangements, corresponding to the six possible starting points as we proceed counterclockwise around the circle. Since six men can be arranged in a row in $6!$ ways, there are $$\frac{6!}{6} = 5!$$ distinguishable circular arrangements of the men.
More generally, we can arrange $n \geq 1$ distinct objects in a circle in $$\frac{n!}{n} = (n - 1)!$$ ways since each circular arrangement corresponds to $n$ linear arrangements, one for which each of the $n$ starting points as we proceed counterclockwise around the circle.
Method 2: We consider arrangements relative to the first man we sit at the table.
We seat man A. We use him as our reference point. As we proceed counterclockwise around the circle, we can seat the remaining five men in $5!$ orders relative to man A.
More generally, given a set of $n$ distinct objects, we can arrange them in $(n - 1)!$ orders as we proceed counterclockwise around the circle relative to the first object we place on the circle.
We can seat the men in $5!$ distinguishable ways. This create six spaces, one to the immediate right of each man. To ensure that no two of the women sit in adjacent seats, we must choose five of those six spaces in which to insert a woman. Once those spaces have been selected, we can arrange the five women in those spaces in $5!$ orders. Hence, the number of possible seating arrangements is $$5! \binom{6}{5} \cdot 5! = 5!P(6, 5)$$