Using $P(A'|B')$ to find $P(A\cup B)$ We have $P(B) = \frac 3 5$ and $P(A'|B') = \frac 1 3$.
I started with $P(A\cap B)' = P(B)' \cdot P(A'|B') = \frac 2 5 \cdot \frac 1 3 = \frac 2 {15}$.
Then converted $P(A\cap B)'$ to $P(A'\cup B')$.
Is $P(A'\cup B') = P(A\cup B)'$, meaning we get $P(A\cup B) = 1 - \frac 2 {15} = \frac {13} {15}$?

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