Given: P(X Y) = (11)/(15) and P(X) = (1)/(3).

For mutually exclusive events, the probability of their union is the sum of their individual probabilities, as $P(X \cap Y) = 0$.

$$P(X \cup Y) = P(X) + P(Y)$$

$$P(Y) = \frac{11}{15} - \frac{1}{3}$$

To subtract, find a common denominator, which is 15.

$$P(Y) = \frac{11}{15} - \frac{1 \times 5}{3 \times 5}$$ $$P(Y) = \frac{11}{15} - \frac{5}{15}$$ $$P(Y) = \frac{11 - 5}{15}$$ $$P(Y) = \frac{6}{15}$$

Simplify the fraction:

$$P(Y) = \frac{2}{5}$$

The probability $P(Y)$ for mutually exclusive events is $\boxed{\frac{2}{5}}$.

For independent events, the probability of their intersection is the product of their individual probabilities: $P(X \cap Y) = P(X) \cdot P(Y)$.
The general addition rule for probabilities is:

$$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$$

Combine terms involving $P(Y)$:

$$\frac{11}{15} - \frac{1}{3} = P(Y) - \frac{1}{3}P(Y)$$ $$\frac{11}{15} - \frac{5}{15} = P(Y)\left(1 - \frac{1}{3}\right)$$ $$\frac{6}{15} = P(Y)\left(\frac{3}{3} - \frac{1}{3}\right)$$ $$\frac{2}{5} = P(Y)\left(\frac{2}{3}\right)$$

$$P(Y) = \frac{2}{5} \div \frac{2}{3}$$ $$P(Y) = \frac{2}{5} \times \frac{3}{2}$$ $$P(Y) = \frac{3}{5}$$

The probability $P(Y)$ for independent events is $\boxed{\frac{3}{5}}$.