You have 12 choices for the first student chosen, 11 choices for the next, then 10, then 9. However, this over-counts everything by a factor of 4! (the number of ways in which four objects can be arranged with regard to order).
Thus, the answer is
$$\frac{12\cdot 11 \cdot 10 \cdot 9}{4!} = 495$$
2.
Let's look at the number of ways that a student can get $x$ answers correct. Clearly, they can get all 8 answers correct in only 1 way. To determine the number of ways a student can get 7 correct you can choose the 7 they get, then choose 1 from the remaining 22: ${8\choose 7}{22\choose 1}$. For the number of ways for a student to get 6 correct it's ${8\choose 6}{22\choose 2}$, etc. So if $X$ is a random variable representing how many answers a student got right,
$$p_X(x)={{8\choose x}{22\choose 8-x}\over{30\choose 8}}$$
Which is just the hypergeometric distribution with $K=8,N=30,$ and $n=8$. So the mean is $$n\frac{K}N=\frac{64}{30}$$and the variance is $$n\frac{K(N-K)(N-n)}{N^2(N-1)}=8\frac{8(30-8)(30-8)}{30^2(30-1)}=\frac{30976}{26100}$$
3.
There may be 4, 5, 6, 7, or 8 out of the 20 answers that are correct. I'm going to assume that, whatever the case, the student still gets 8 choices. The method I would use to determine the mean and variance of the number of correct responses made by a student would be to make a probability mass function for each of the five possible quiz designs (by which I mean quizzes with 8, 7, 6, 5, or 4 distinct correct responses). Let's call these functions $p_i(x)$, where $i$ is the number of distinct correct responses in the quiz. Then, assuming each of the 5 quiz designs is equally likely, the total probability mass function would be
$$p_X(x)=\frac15p_8(x)+\frac15p_7(x)+\frac15p_6(x)+\frac15p_5(x)+\frac15p_4(x)$$
If the 5 quiz designs aren't equally likely you could of course adjust the weights on the $p_i(x)$'s to reflect that. With this function found you could find the mean and variance as you usually do for discrete random variables.
It looks like quite a task to completely find each of the $p_i(x)$'s, so instead of doing that I'll find a few of the values from a couple of them to illustrate the logic.$$\\$$
$\mathbf{p_8(x)}$:
If there were 8 distinct correct answers, then this would be hypergeometrically distributed as above but with $N=20$.
$\mathbf{p_7(x)}$:
If there were 7 distinct correct answers, then one of them would be the correct response to 2 of the questions. Let's call that one the 'double' response and the others 'single' responses. The number of ways of getting all 8 questions correct is the number of ways of choosing all 7 of these in 8 guesses, $$p_7(8)={{7\choose 7}{13\choose 1}\over {20\choose8}}$$A student can only score 7 out of 8 by getting the 'double' response and 5 of the others, so $$p_7(7)={{1\choose 1}{6\choose 5}{13\choose 2}\over{20\choose8}}$$A student can score 6 out of 8 by getting all six of 'singles' or by getting the 'double' and four of the 'single' answers, which means $$p_7(6)={{1\choose 0}{6\choose6}{13\choose2}+{1\choose1}{6\choose4}{13\choose3}\over{20\choose 8}}$$
$\mathbf{p_6(x)}$:If there were 6 distinct correct answers, then two of them would be 'double' solutions and four would be 'single' solutions. To get 4 out of the 8 questions correct a student could get none of the double solutions and all four of the singles, one double and two singles, or both doubles, so $$p_6(4)={{2\choose 0}{4\choose 4}{14\choose 4}+{2\choose 1}{4\choose 2}{14\choose 5}+{2\choose 2}{4\choose 0}{14\choose 6}\over{20\choose8}}$$
Similar logic should get you all the values for all the $p_i(x)$'s. And, as said above, with those found you would be able to calculate the mean and variance.
Best Answer
It would be $8!$ if it were a line of people, but a circle means there's no defined "start point" like there is with the line.
Since there are $8$ ways to "start" the circle, there are $\frac{8!}{8} = 7!$ ways to create a unique circle of people in terms of relative ordering (i.e. where rotations do not matter).
See also: MathWorld: Circular Permutation