By inclusion/exclusion,
[tex]n(P' \cup Q) = n(P') + n(Q) - n(P' \cap Q)[/tex]
We have
[tex]n(\xi) = n(P) + n(P') \implies n(P') = 28 - n(P)[/tex]
so that
[tex]n(P' \cup Q) = (28 - n(P)) + n(Q) - 2n(P) = 28 - 3n(P) + n(Q)[/tex]
Now,
[tex]n(\xi) = 28 \implies n(P \cup Q) = 28 - 7 = 21[/tex]
and by inclusion/exclusion,
[tex]n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)[/tex]
Decompose [tex]Q[/tex] into the union of two disjoint sets:
[tex]Q = (P \cap Q) \cup (P' \cap Q)[/tex]
Since they're disjoint,
[tex]n(Q) = n(P\cap Q) + n(P'\cap Q) \implies n(Q) = n(P\cap Q) + 2n(P)[/tex]
[tex]\implies n(P \cup Q) = n(P) + (n(P\cap Q) + 2n(P)) - n(P \cap Q)[/tex]
[tex]\implies 21 = 3n(P)[/tex]
[tex]\implies n(P) = 7[/tex]
From the Venn diagram, we see there are 3 elements unique to [tex]P[/tex] - by the way, this is the set [tex]P \cap Q'[/tex] - so [tex]n(P\cap Q) = 7-3 = 4[/tex], and it follows that
[tex]n(Q) = n(P\cap Q) + 2n(P) = 4 + 2\times7 = 18[/tex]
Finally, we get for (a)
[tex]n(P' \cup Q) = 28 - 3n(P) + n(Q) = 28 - 3\times7 + 18 = \boxed{25}[/tex]
For (b), we have by inclusion/exclusion that
[tex]n(P \cup Q') = n(P) + n(Q) - n(P \cap Q') = 7 + 18 - 3 = \boxed{22}[/tex]
