1. Given data

We are given: $P(X)=\frac{1}{3}, \quad P(X\mid Y)=\frac{1}{2}, \quad P(Y\mid X)=\frac{2}{5}.$

We use the conditional probability formulas: $P(X\mid Y)=\frac{P(X\cap Y)}{P(Y)}, \qquad P(Y\mid X)=\frac{P(X\cap Y)}{P(X)}.$

2. Find $P(X\cap Y)$ using $P(Y\mid X)$

Since $P(Y\mid X)=\frac{P(X\cap Y)}{P(X)}=\frac{2}{5},$ we get $P(X\cap Y)=P(Y\mid X)\cdot P(X)=\frac{2}{5}\cdot \frac{1}{3}=\frac{2}{15}.$

3. Find $P(Y)$ using $P(X\mid Y)$

Now, $P(X\mid Y)=\frac{P(X\cap Y)}{P(Y)}=\frac{1}{2}.$ So, $\frac{2}{15\,P(Y)}=\frac{1}{2}$ or directly, $P(Y)=\frac{P(X\cap Y)}{P(X\mid Y)}=\frac{\frac{2}{15}}{\frac{1}{2}}=\frac{4}{15}.$

So option A is correct.

4. Check option B: $P(X'\mid Y)=\frac{1}{2}$

Using complement inside event $Y$: $P(X'\mid Y)=1-P(X\mid Y)=1-\frac{1}{2}=\frac{1}{2}.$

So option B is correct.

5. Check option C: $P(X\cup Y)=\frac{2}{5}$

Use $P(X\cup Y)=P(X)+P(Y)-P(X\cap Y).$ Substitute values: $P(X\cup Y)=\frac{1}{3}+\frac{4}{15}-\frac{2}{15}.$ Convert to common denominator: $\frac{1}{3}=\frac{5}{15}.$ Hence, $P(X\cup Y)=\frac{5}{15}+\frac{4}{15}-\frac{2}{15}=\frac{7}{15}.$ But $\frac{2}{5}=\frac{6}{15}.$ Since $\frac{7}{15}\ne \frac{2}{5},$ option C is false.

6. Check option D: $P(X\cap Y)=\frac{1}{5}$

We already found $P(X\cap Y)=\frac{2}{15}.$ Now, $\frac{1}{5}=\frac{3}{15}.$ Since $\frac{2}{15}\ne \frac{1}{5},$ option D is false.

7. Final answer

Correct options are: $\boxed{A,\ B}$

This matches the stored correct answer.