Question 1
You're right, it is $1/C^Q$.
Question 3
By the law of large numbers, the average of the results obtained from a large number of trials should be close to the expected value.
The expected value, from the other side, is obviously $\frac{1}{C}$ (as, for each question, there is $\frac{1}{C}$ probability of getting $1$ and $\frac{C-1}{C}$ probability of getting $0$).
Question 4
Changing $C$ directly changes the expected average score. The number of questions and students affects the dispersion of the "average score".
Question 5
Results may vary considerably across tests, although more the number of questions and students is, the less likely considerable differences are.
For example, on the first test all students may accidentally guess all answers (which is possible, although unlikely); on the second test it is possible that none of the students will guess the right answer.
Question 6
As long as the correct answers are distributed uniformly, it doesn't matter which answer will you choose on a specific question, the probability of guessing is $\frac{1}{C}$. And, of course, the probability of getting the right answer for $n+1$-th question does not depend on what answer was chosen for $n$-th question. Both strategies are equivalent.
The results you got in your test run may be explained by e.g. non-uniform correct answers distributions (e.g. the second answer is correct for 50% of questions, while the first and the third answers are correct for 25% of questions) and checking against some unlucky answer (e.g. the first one). In such a case, choosing a random answer will give you 33.3% expected score, while always choosing the first answer will give you only 25% of expected score.
The idea is to construct a probability distribution on the 10 K-Prim type questions. The probability distribution for a single question is $$\begin{align*} p_0 = \Pr[S = 0] &= \sum_{k=0}^2 \binom{4}{k} (0.5)^k (0.5)^{4-k} = \frac{11}{16}, \\ p_1 = \Pr[S = 0.5] &= \binom{4}{3} (0.5)^3 (0.5)^1 = \frac{1}{4}, \\ p_2 = \Pr[S = 1] &= \binom{4}{4} (0.5)^4 (0.5)^0 = \frac{1}{16}. \end{align*} $$ This assumes that for each such question, the choice of True/False is equally likely for each of the four answers, and the each answer is independent, thus the number of correct answers follows a ${\rm Binomial}(4,0.5)$ distribution.
Next, the distribution of the sum of the scores of 10 K-Prim questions can be derived from the multinomial distribution, though it is somewhat tedious to compute: let $X_0$, $X_1$, $X_2$ be random variables that count the number of $0$-point, $0.5$-point, and $1$-point scores out of the 10 questions. Then $$\Pr[(X_0, X_1, X_2) = (a,b,c)] = \frac{10!}{a! b! c!} p_0^{a} p_1^b p_2^c.$$ Then we can tabulate the sum; we do this in Mathematica:
Flatten[Table[{b/2 + c, PDF[MultinomialDistribution[10, {11/16, 1/4, 1/16}],
{10 - b - c, b, c}]}, {b, 0, 10}, {c, 0, 10 - b}], 1]
Table[{k, Total[Select[%, #[[1]] == k &]][[2]]}, {k, 0, 10, 1/2}]
which gives us the desired probability distribution for these 10 questions. Call this random variable $K$. Now, for the remaining 20 questions, the total point count is simple; it is simply $A \sim {\rm Binomial}(20, 0.2)$. So the probability that the total score is at least $18$ out of $30$ is $$\sum_{k=0}^{20} \Pr[K = k/2]\Pr[A \ge 18 - k/2].$$ Again, we use Mathematica:
Sum[%[[k, 2]] (1 - CDF[BinomialDistribution[20, 1/5], 18 - k/2]),
{k, 1, Length[%]}]
This gives us $$\frac{8327843221553613}{2^9 \cdot 10^{20}} \approx 1.62653 \times 10^{-7}.$$ This is so small that it is unlikely that a naive simulation approach will be able to approximate it.
Best Answer
Very small. To get at most 7 right, $$ \sum_{k= 0}^7 {100\choose k}{1\over 2^{100}}.$$ This is a wee-tiny number.