Let's analyze the given conditions and evaluate which options are correct.

Given:

1. $$ P(X|Y) = \frac{1}{2} $$


2. $$ P(Y|X) = \frac{1}{3} $$


3. $$ P(X \cap Y) = \frac{1}{6} $$

Now, let's proceed step by step through each option.

Option A: $$ P(X \cup Y) = \frac{2}{3} $$

We can use the formula for the union of two events:

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

To find $$ P(X) $$ and $$ P(Y) $$, we use the definitions of conditional probability:

$$ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} $$

$$ \frac{1}{2} = \frac{\frac{1}{6}}{P(Y)} $$

$$ P(Y) = \frac{1}{6} \times 2 = \frac{1}{3} $$

Similarly,

$$ P(Y|X) = \frac{P(X \cap Y)}{P(X)} $$

$$ \frac{1}{3} = \frac{\frac{1}{6}}{P(X)} $$

$$ P(X) = \frac{1}{6} \times 3 = \frac{1}{2} $$

Now, substituting these values into the union formula:

$$ P(X \cup Y) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} $$

To add these fractions, we need a common denominator. The common denominator is 6:

$$ P(X \cup Y) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3} $$

Hence, Option A is correct.

Option B: $$ X $$ and $$ Y $$ are independent

For two events $$ X $$ and $$ Y $$ to be independent, the following condition should hold:

$$ P(X \cap Y) = P(X) \times P(Y) $$

We already found that:

$$ P(X) = \frac{1}{2} $$

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

$$ P(X \cap Y) = \frac{1}{6} $$

Checking the independence condition:

$$ P(X) \times P(Y) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} $$

Since this is equal to $$ P(X \cap Y) $$, $$ X $$ and $$ Y $$ are indeed independent.

Hence, Option B is correct.

Option C: $$ X $$ and $$ Y $$ are not independent

From the previous analysis, we have shown that $$ X $$ and $$ Y $$ are independent.

Hence, Option C is not correct.

Option D: $$ P((X^c) \cap Y) = \frac{1}{3} $$

To find $$ P((X^c) \cap Y) $$, we use the fact that:

$$ P(Y) = P((X \cap Y) \cup (X^c \cap Y)) $$

Since $$ (X \cap Y) $$ and $$ (X^c \cap Y) $$ are disjoint events, we can write:

$$ P(Y) = P(X \cap Y) + P(X^c \cap Y) $$

Given:

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

$$ P(X \cap Y) = \frac{1}{6} $$

Substituting these,

$$ \frac{1}{3} = \frac{1}{6} + P(X^c \cap Y) $$

Solving for $$ P(X^c \cap Y) $$:

$$ P(X^c \cap Y) = \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6} $$

This contradicts Option D, as it says $$ \frac{1}{3} $$.

Hence, Option D is not correct.

In conclusion:

Option A and Option B are correct.

Option C and Option D are not correct.