This is an interesting problem because, at first, it doesn't look like there is enough information. Most notably, we don't have P(B), so we can't use the formula P(A∪B) = P(A) + P(B) - P(A∩B). However, we can get P(A'), and therefore P(B∩A').
If P(A) = 1/3, then P(A') = 1 - 1/3 = 2/3.
Then by using the formula P(A∩B) = P(B|A')P(A'), P(A∩B') = 1/4 · 2/3 = 1/6.
How does that help? Consider how this would look on a Venn diagram (which, unfortunately, I can't display here). If we shade the circle for A, that's represented by P(A), which is 1/3. After that, what else needs to be shaded to get A∪B? Not all of circle B, but just the part that is outside circle A. That's precisely what B∩A' is.
So all we need now is to add P(A) and P(A∩B'). 1/3 + 1/6 = 1/2.
