Let $$E_1$$ and $$E_2$$ be two events such that the conditional probabilities $$P(E_1 | E_2) = \frac{1}{2}$$, $$P(E_2 | E_1) = \frac{3}{4}$$ and $$P(E_1 \cap E_2) = \frac{1}{8}$$. Then

We are given: $$P(E_1 | E_2) = \frac{1}{2}$$, $$P(E_2 | E_1) = \frac{3}{4}$$, and $$P(E_1 \cap E_2) = \frac{1}{8}$$.

We use the relation $$ P(E_1 \cap E_2) = P(E_1 | E_2) \cdot P(E_2) \implies \frac{1}{8} = \frac{1}{2} \cdot P(E_2) \implies P(E_2) = \frac{1}{4} $$ and similarly apply $$ P(E_1 \cap E_2) = P(E_2 | E_1) \cdot P(E_1) \implies \frac{1}{8} = \frac{3}{4} \cdot P(E_1) \implies P(E_1) = \frac{1}{6} $$.

Next, we check each option.

Option A asks whether $$P(E_1 \cap E_2) = P(E_1) \cdot P(E_2)$$. The left-hand side is $$\frac{1}{8}$$, while the right-hand side is $$\frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$$, so they are not equal.

Option B considers whether $$P(E_1' \cap E_2') = P(E_1') \cdot P(E_2)$$. We note that
$$P(E_1' \cap E_2') = 1 - P(E_1 \cup E_2) = 1 - (P(E_1) + P(E_2) - P(E_1 \cap E_2))$$
$$= 1 - \frac{1}{6} - \frac{1}{4} + \frac{1}{8} = 1 - \frac{4 + 6 - 3}{24} = 1 - \frac{7}{24} = \frac{17}{24}$$
and
$$P(E_1') \cdot P(E_2) = \frac{5}{6} \cdot \frac{1}{4} = \frac{5}{24}$$, which are not equal.

Option C checks whether $$P(E_1 \cap E_2') = P(E_1) \cdot P(E_2)$$. We compute
$$P(E_1 \cap E_2') = P(E_1) - P(E_1 \cap E_2) = \frac{1}{6} - \frac{1}{8} = \frac{4-3}{24} = \frac{1}{24}$$
and
$$P(E_1) \cdot P(E_2) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$$, showing that they are equal.

Option D asks if $$P(E_1 \cup E_2) = P(E_1) \cdot P(E_2)$$. Since $$P(E_1 \cup E_2) = \frac{7}{24}$$ and $$P(E_1) \cdot P(E_2) = \frac{1}{24}$$, they do not match.

Option C: $$P(E_1 \cap E_2') = P(E_1) \cdot P(E_2)$$.