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.
Your answer to $1$.) is right, with the right approach for this sort of problem.
Your answer to $2$.) is also almost right, except you got the doctor's gender wrong :-)
Best Answer
For $1$, each family forms a block. There are $7!/7$ inequivalent ways to arrange the $7$ men, women and blocks. With the single women also forming a block, there are $6!/6$ inequivalent ways. In the first case, there are $3!^2$ internal permutations for the families, and in the second case there are $3!^2$ internal permutations for the families and $2$ for the single women. Thus the total number of inequivalent admissible arrangements is
$$ (3!^2)7!/7-2(3!^2)6!/6=17280\;. $$
If the seats are numbered and each child sits between their parents, there are $11\cdot7!/7$ arrangements of the men, women and family blocks, and $2^2$ internal arrangements for the two families, yielding
$$ 2^2\cdot11\cdot7!/7=31680 $$
different arrangements.