Using $P(A'|B')$ to find $P(A\cup B)$
You have made two mistakes, but got the right answer! $P(A'\cap B')=P(A'|B') P(B')=\frac 1 3 (1-\frac 3 5)= \frac 2 {15}$. $P(A\cup B)=1-P(A' \cap B')=1-\frac 2 {15}=\frac {13} {15}$.
$\frac{1}{3}=P(A^c/B^c)=\frac{P(A^c \cap B^c)}{P(B^c)}=\frac{P((A \cup B)^c)}{1-P(B)}=\frac{1-P(A \cup B)}{1-\frac{3}{5}}$ $$\implies \frac{1×2}{3×5}=1-P(A \cup B)$$ $$\implies P(A \cup B)=\frac{13}{15}$$