All of your answers are correct. I will assume that only the relative order of the people matters, that the men are named Barney, David, and Fred, and that the women are named Ashley, Carly, and Emma.
What is the probability that all three men will be seated next to each other?
We seat Ashley. The remaining people can be seated in $5!$ ways as we proceed clockwise around the table.
For the favorable cases, we have two blocks of people to arrange. Since the blocks of men and women must be adjacent, this can be done in one way. The block of men can be arranged in $3!$ ways. The block of women can be arranged in $3!$ ways. Hence, the probability that all the men are seated next to each other is
$$\frac{3!3!}{5!} = \frac{3}{10}$$
What is the probability that starting at the northernmost seat and going clockwise around the table, a second man will be encountered before a second woman?
You made good use of symmetry. Nice solution.
What is the probability that no two men are seated directly opposite each other?
We count arrangements in which two men are seated directly opposite each other. There are $\binom{3}{2}$ ways to select the men. Once they are seated, there are $4$ ways to seat the third man relative to the already seated man whose name appears first alphabetically and $3!$ ways to seat the women as we proceed clockwise around the table from the third man. Hence, there are
$$\binom{3}{2}\binom{4}{1}3!$$
seating arrangements in which two men are opposite each other. Hence, the probability that no two men are opposite each other is
$$1 - \frac{\binom{3}{2}\binom{4}{1}3!}{5!} = 1 - \frac{3}{5} = \frac{2}{5}$$
Suppose that the six people are named Ashley, Barney, Carly, David, Emma, and Fred. What is the probability that starting with Ashley and going clockwise, the six people are seated in alphabetical order.
There is only one permissible seating arrangement. Hence, the probability that they are seated in alphabetical order is
$$\frac{1}{5!} = \frac{1}{120}$$
Your first approach is correct.
The way to get an answer to this is to look at the problem inductively. You can pair the first woman off with 10 men. The second woman could then choose from 9 men. Going on like this you would conclude that the tenth woman could choose from 1 man. Hence your answer is going to be $10 \times 9 \times \dots \times 1 = 10!$.
For the second approach, if $N=4$ ($2$ men, $2$ woman) then the first woman can pair with 2 men while the second woman can pair with 1 man. Therefore, it would be $2!$.
As noted in the comments, for $2$ men and $2$ woman there are only $2$ possibilities
$$m_1w_1, m_2w_2\tag{1}$$
$$m_2w_1, m_1w_2\tag{2}$$
which is why it is $2$!.
If $N=6$ ($3$ men, $3$ woman) then the first woman can pair with 3 men. The second woman can pair with 2 men and the third woman can pair with one man. Therefore, it would be $3!$. This can be seen by
$$m_1w_1, m_2w_2, m_3w_3\tag{1}$$
$$m_2w_1, m_1w_2, m_3w_3\tag{2}$$
$$m_3w_1, m_2w_2, m_1w_3\tag{3}$$
$$m_1w_2, m_2w_3, m_3w_1\tag{4}$$
$$m_3w_2, m_2w_1, m_1w_3\tag{5}$$
$$m_2w_3, m_1w_1, m_3w_2\tag{6}$$
Best Answer
Let's say that the positions are numbered like this:
$1|2|3|4$
$5|6|7|8$
That means that the seat $5$ is opposite of seat $1$, seat $6$ is opposite of seat $2$, etc.
You can chose between $8$ people for whom to put in seat number $1$.
After this, you can chose between $4$ people for whom to put opposite of seat number $1$(that is in seat number $5$).
Next you can chose between $6$ people for whom to put in seat number $2$, and then you can chose between $3$ people to put in seat number $6$
Continuing this way we can find that there are $8*4*6*3*4*2*2*1$ ways to arrange the people the way you want.