Suppose that 3 men and 3 women will be randomly seated at a round table having 6 evenly-spaced seats Suppose that $3$ men and $3$ women will be randomly seated at a round table having $6$ evenly-spaced seats (with each seat being directly opposite one other seat).
(a) What is the probability that all $3$ men will be seated next to one another?
(b) What is the probability that starting at the northernmost seat and going clockwise around the table, a 2nd man will be encountered before a 2nd woman?
(c) What is the probability that no two men are seated directly opposite of one another?
(d) Suppose that the $6$ people are named Ashley, Barney, Carly, David, Emma, and Fred. What is the probability that starting with Ashley and going clockwise, the 6 people are seated in alphabetical order?
Attempted Solution:
(a) There are $6!$ = $720$ ways to seat these $6$ people. There are 6 different ways you can seat all three men next to each other, if you consider them as a clump. Each clump of three has $3!$ = 6 different ways to be arranged. Thus, there is $6*3!*3!$ = $216$ different arrangements. This gives a probability of $216\over{720}$ = $.3$.
(b) I wasn't sure about this but I realized that in any situation, either a 2nd man is reached before a 2nd woman or a 2nd woman is reached before a 2nd man. This gives a probability of $.5$.
(c) After drawing a picture of the table, this appears to be the same situation as part (a), giving a probability of $.3.$ I just realized this assumption was incorrect. Now I am getting that there are 8 different arrangements when considering males and females as clumps, with $3!$ different ways of arranging each clump, giving $8*3!*3!$ = $288$. This gives a probability of $288\over{720}$ = $.4$.
(d) $6\over{6}$ * $1\over{5}$ * $1\over{4}$ * $1\over{3}$ * $1\over{2}$ * $1\over{1}$ = $.00833$.
Any corrections to my attempted solutions would be greatly appreciated.

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 
$$\frac{1}{5!} = \frac{1}{120}$$
