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
$$\frac{1}{5!} = \frac{1}{120}$$
You can just use the first brother to orient the table and do everything from there. Seating him prohibits seven seats to the other brother, the one he is in and three on either side. Seat the second brother in one of $13$ places, then the rest of the people in $18!$ ways, for a total of $13 \cdot 18!$
Your multiplying by $2$ double counts each arrangement, once when you seat each brother first.
Best Answer
The book answer is correct
There is the master, mistress and $12$ guests, and in the absence of information to the contrary, it is customary to take chairs to be unnumbered, so there is only one way to seat the hosts opposite to each other.
In effect, the remaining $12$ seats become numbered with respect to the mistress at the $12$ o'clock position, say, and the master at $6$ o'clock position.
In between there are $6$ seats in each half, but the two specified people can only sit together in $2\cdot5$ ways in each half, i.e. $2\cdot10$ ways altogether.
The remaining $10$ guests can then be arranged in $10!$ ways
Putting everything together, ans $= 2*10*10!$