Notice that
[tex](A\cup B)\cap(A\cup B')=A[/tex]
So
[tex]\mathbb P\bigg((A\cup B)\cap(A\cup B')\bigg)=\mathbb P(A)[/tex]
Recall the inclusion/exclusion principle:
[tex]\mathbb P\bigg((A\cup B)\cap(A\cup B')\bigg)=\mathbb P(A\cup B)+\mathbb P(A\cup B')-\mathbb P\bigg((A\cup B)\cup(A\cup B')\bigg)[/tex]
As [tex](A\cup B)\cup(A\cup B')=U[/tex] (the universal set), we have
[tex]\mathbb P(A)=\mathbb P(A\cup B)+\mathbb P(A\cup B')-\mathbb P(U)[/tex]
[tex]\implies\mathbb P(A)=0.76+0.87-1[/tex]
[tex]\implies\mathbb P(A)=0.63[/tex]
