For part a $X_i$: score on question $i$.
$P(X_i=1)=p_i+\frac{1-p_i}{5}=\frac{1}{5}+\frac{4}{5}p_i$
$P(X_i=-\frac{1}{4})=(1-p_i)*\frac{4}{5}=\frac{4}{5}-\frac{4}{5}p_i$
So
$\mu=\Sigma(\frac{1}{5}+\frac{4}{5}p_i-{1\over4}(\frac{4}{5}-\frac{4}{5}p_i))=\Sigma p_i$
And
$\sigma=\Sigma((\frac{1}{5}+\frac{4}{5}p_i)(1)^2+(\frac{4}{5}-\frac{4}{5}p_i)(-\frac{1}{4})^2-p_i^2)$
$\sigma=\Sigma(1-p_i)(\frac{1}{4}+p_i)$
Which is not the result given in the question - is the question wrong?
To maximize skill penetration, award 1 point for each correct answer, and -1000 for each incorrect answer. Note: this is not a joke, this is how to only get answers that the student is entirely confident of.
In your example, the student's knowledge is net zero, getting the same score as random guessing. Such a student is unlikely to adapt his or her test-taking strategy to subtle rewards and penalties.
Additional commentary regarding gambling: Suppose you let each student give a weight to each question, of any real number in $[0,1]$, as well as the T/F answer. If right, the student gains that many points, if wrong, loses. Naively one might think that this would encourage students to weigh questions more if they are more confident, and less otherwise. In fact this is not true. If I am 51% confident that the answer is true, I will maximize my score by weighing it 1 rather than anything else.
Suppose instead you allow students to weight as above, but this time if they are wrong they lose DOUBLE their wager. If I am right with probability $p$ my expected score is $pw-2(1-p)w=w(3p-2)$. All this change does is move the threshhold from $p=\frac{1}{2}$ to $p=\frac{2}{3}$. If I believe that $p>\frac{2}{3}$, I should bet the maximum, otherwise I should bet 0.
Result: under all similar gambling schemes, there is never any reason to bet anything other than the maximum (or zero/leave the question blank).
Best Answer
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 $\sqrt{20}$.