a.
[tex]B\cap C=(B\cap C)\cap(A\cup A^c)=(A\cap B\cap C)\cup(A^c\cap B\cap C)[/tex]
which means
[tex]n(B\cap C)=n(A\cap B\cap C)+n(A^c\cap B\cap C)[/tex]
[tex]\implies n(A^c\cap B\cap C)=28-n(A\cap B\cap C)[/tex]
Similarly,
[tex]n(A\cap B)=n(A\cap B\cap C)+n(A\cap B\cap C^c)[/tex]
[tex]\implies n(A\cap B\cap C)=27-12=15[/tex]
Then
[tex]n(A^c\cap B\cap C)=28-15=\boxed{13}[/tex]
b.
[tex]n(A)=n(A\cap(B\cup C))+n(A\cap(B\cup C)^c)[/tex]
[tex]n(A)=n(A\cap(B\cup C))+n(A\cap B^c\cap C^c)[/tex]
[tex]\implies n(A\cap B^c\cap C^c)=39-n(A\cap(B\cup C))[/tex]
Note that [tex]A\cap(B\cup C)=(A\cap B)\cup(A\cap C)[/tex].
By the inclusion-exclusion principle,
[tex]n(A\cap(B\cup C))=n(A\cap B)+n(A\cap C)-n((A\cap B)\cap(A\cap C))[/tex]
and [tex](A\cap B)\cap(A\cap C)=A\cap B\cap C[/tex], so
[tex]n(A\cap(B\cup C))=27+21-15=33[/tex]
Then
[tex]n(A\cap B^c\cap C^c)=39-33=\boxed6[/tex]
