Let $X_i=1$ if the student is right on the $i$-th question, and $0$ if she is not right. Let $M_i$ be the student's mark on the $i$-th question. Then $M_i=2X_i-1$. The test score $S$ is then given by
$$S=M_1+M_2+\cdots+M_{20}.$$
It is standard that the standard error of $X_i$ is $\sqrt{(1/2)(1-1/2)}=1/2$. So the standard error of $2X_i-1$ is $(2)(1/2)=1$.
The standard error of the sum $M_1+M_2+\cdots+M_{20}$ is therefore $(\sqrt{20})(1)$.
Remark: when you divided the (not quite right) standard error per question by $\sqrt{20}$, you were finding the standard error of the average mark per question, not the standard error of the total mark.
We could also find the SE of each $M_i$ directly. The variance of $M_i$ is $E(M_i^2)-(\mu_i)^2$, where $\mu_i$ is the mean of $M_i$. But the mean of $M_i$ is $0$. And $M_i^2=1$ always, so $E(M_i^2)=1$. It follows that the variance of $M_i$ is $1$, and therefore the SE of $M_i$ is $1$.
The $M_i$ are assumed independent. The total mark is the sum of the $M_i$, so has variance $(20)(1)$, and therefore standard deviation $\sqrt{20}$.
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
Perhaps easier to compute the fail case. Will fail if answers (out of the remaining 8) 0 or 1 questions.
$ P(\text{Fail}) = {8 \choose 0} (\frac12)^8 + {8 \choose 1}(\frac12)^8 = \frac{1+8}{256} $