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.
Let's mark one seat as "Seat $1$" and then number the seats, incrementing by $1$ as we go counter clockwise and stopping when we get to Seat $1$ again. Seat $1$ can either be a man or a woman, which is $2$ possibilities. Then, all of the odd seats must have the same gender as Seat $1$ and all of the even seats must have the opposite gender as Seat $1$. However, we can rearrange the people in the odd seats and since there are $5$ odd seats ($1, 3, 5, 7, 9$), this has $5!$ possibilities. Similarly, we can rearrange the people in the even seats and since there are $5$ even seats ($2, 4, 6, 8, 10$), this also has $5!$ possibilities.
Therefore, we multiply all of the possibilities together to find that there are $2*5!*5!$ ways to arrange all of the people so that no two adjacent people have the same gender.
Best Answer
The extra factor of two comes from which seats the women actually sit on. For example, if you number the seats clockwise from 1 to 16, your solution gives all cases for which the women sit on odd seats and the men sit on even seats. But the other situation is where women sit on even seats and men sit on odd seats, so there are actually twice as many solutions.