How to solve this problem using set theory?

Let $e$ be the number of students who passed English but not math,
$m$ the number of students who passed math but not English,
$a$ the number of students who passed both tests, $z$ the number of students who passed neither test.

$$36=e+m+a+z$$ $$25=e+a$$ $$28=m+a$$ $$20=a$$

$4$ equations in $4$ unknowns!

Spoiler: $$a=20$$ $$e=(e+a)-a=25-20=5$$ $$m=(m+a)-a=28-20=8$$ $$e+m+a+z=36$$ $$5+8+20+z=36$$ $$33+z=36$$ $$z=36-33=3$$

Denote by $S$ the set of all students, by $E$ the set of all students that passed English and by $M$ the set of all students that passed Math. Then what you know is: $|S|=36$, $|E|=25$, $|M|=28$ and $|Ecap M|=20$. Then what you are looking for is:
a. $|(Ssetminus E)cap (Ssetminus M)|=$
b. $|Esetminus M|=$
c. $|Msetminus E|=$
So you can use De Morgan's laws and Martin Sleziak's answer from here on to solve...

If $E$ is the set of the students that passed English and $M$ are those that passed Math then you have:

$|E|=25$, $|M|=28$, $|Ecap M|=20$

From the formula

$$|Ecup M|=|E|+|M|-|Ecap M|$$

you get that $|Ecup M|=33$, which means that 33 students passed at least one subject.

I guess you can go on from here...

If you prefer diagrams instead of formulas, you can try to draw something like the diagrams shown here (just with different numbers): Venn Diagrams: Exercises (Purplemath). To prevent the link rot, here is also Wayback Machine link.

Perhaps I should have also added the link to the Wikipedia article Inclusion–exclusion principle.