Let A and B be two events such that $$P(B|A) = \frac{2}{5}$$, $$P(A|B) = \frac{1}{7}$$ and $$P(A \cap B) = \frac{1}{9}$$. Consider $$(S_1): P(A' \cup B) = \frac{5}{6}$$, $$(S_2): P(A' \cap B') = \frac{1}{18}$$. Then

We are given $$P(B|A) = \frac{2}{5}$$, $$P(A|B) = \frac{1}{7}$$, and $$P(A \cap B) = \frac{1}{9}$$.

From $$P(B|A) = \frac{P(A \cap B)}{P(A)}$$, we get $$P(A) = \frac{P(A \cap B)}{P(B|A)} = \frac{1/9}{2/5} = \frac{5}{18}$$.

From $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$, we get $$P(B) = \frac{P(A \cap B)}{P(A|B)} = \frac{1/9}{1/7} = \frac{7}{9}$$.

Now we check statement $$(S_1)$$: $$P(A' \cup B) = \frac{5}{6}$$.

Using $$P(A' \cup B) = 1 - P(A \cap B') = 1 - [P(A) - P(A \cap B)]$$:

$$P(A' \cup B) = 1 - \left(\frac{5}{18} - \frac{1}{9}\right) = 1 - \left(\frac{5}{18} - \frac{2}{18}\right) = 1 - \frac{3}{18} = 1 - \frac{1}{6} = \frac{5}{6}$$

So $$(S_1)$$ is true.

Now we check statement $$(S_2)$$: $$P(A' \cap B') = \frac{1}{18}$$.

Using De Morgan's law: $$P(A' \cap B') = 1 - P(A \cup B) = 1 - [P(A) + P(B) - P(A \cap B)]$$:

$$P(A' \cap B') = 1 - \left(\frac{5}{18} + \frac{7}{9} - \frac{1}{9}\right) = 1 - \left(\frac{5}{18} + \frac{6}{9}\right) = 1 - \left(\frac{5}{18} + \frac{12}{18}\right) = 1 - \frac{17}{18} = \frac{1}{18}$$

So $$(S_2)$$ is also true.

Hence, both $$(S_1)$$ and $$(S_2)$$ are true, and the correct answer is Option A.