1. Consult the linked question. [tex]n(A\cap B\cap W) = \boxed{3}[/tex].
2. We have
[tex]n(A\cap B) = n(A\cap B \cap W) + n(A\cap B \cap W') \\\\ \implies 11 = 3 + n(A\cap B \cap W') \\\\ \implies n(A\cap B\cap W') = 8[/tex]
Similarly we can find
[tex]n(A\cap B' \cap W) = 3[/tex]
[tex]n(A'\cap B\cap W) = 6[/tex]
[tex]n(A\cap B'\cap W') = 16[/tex]
[tex]n(A'\cap B\cap W') = 8[/tex]
[tex]n(A'\cap B'\cap W) = 12[/tex]
Then the total number of students that applied to at least two of the universities is
[tex]\underbrace{n(A\cap B\cap W') + n(A\cap B'\cap W) + n(A'\cap B\cap W)}_{\text{only 2}} + \underbrace{n(A\cap B\cap W)}_{\text{all 3}} = \boxed{20}[/tex]
3. There's a small ambiguity here. Are we interested in students that applied to zero universities? If so, there are
[tex]\underbrace{n(U\setminus(A\cup B\cup W))}_{\text{none}} + \underbrace{n(A\cap B'\cap W') + n(A'\cap B\cap W') + n(A'\cap B'\cap W)}_{\text{only 1}} = \boxed{40}[/tex]
If we want students that applied to at least 1 school, we omit the first term and get a total of 36.
4. Split this set of students into those that also applied to Wachemo and those that did not.
[tex]n(A \cap B') = n(A\cap B' \cap W) + n(A\cap B'\cap W') = \boxed{19}[/tex]
