Recover the marginal before testing independence

$P(B\cap A')=P(A')P(B\mid A')=(1-0.7)(0.4)=0.12$. Hence $P(B)=P(B\cap A)+P(B\cap A')=0.28+0.12=0.4$.

$P(B\mid A')=P(B)=0.4$ with $P(A')>0$, so B is independent of A′ and therefore of A. This establishes that A and B are independent.

$P[(A\cap B)\mid(A\cup B)]=P[(A\cap B)\cap(A\cup B)]/P(A\cup B)=P(A\cap B)/P(A\cup B)$. The union is the disjoint combination of A and B outside A: $P(A\cup B)=0.7+0.12=0.82$. Thus $0.28/0.82=14/41\approx0.341$ (3 s.f.).