Inclusion-exclusion tells us that $P(A\cup B) = P(A)+P(B)-P(A\cap B)$
Rearranging, we have $P(A\cap B)=P(A)+P(B)-P(A\cup B)$
We have then $P(A\cap B)=\frac{2}{5}+\frac{5}{7}-P(A\cup B)\geq \frac{2}{5}+\frac{5}{7}-1>0$
Thus, $P(A\cap B)>0$ and so $A\cap B\
eq \emptyset$