For $$a \in \mathbb{C}$$, let $$A = \{z \in \mathbb{C}: \text{Re}(a + \bar{z}) > \text{Im}(\bar{a} + z)\}$$ and $$B = \{z \in \mathbb{C}: \text{Re}(a + \bar{z}) < \text{Im}(\bar{a} + z)\}$$. Then among the two statements:
$$(S1)$$: If $$\text{Re}(a), \text{Im}(a) > 0$$, then the set $$A$$ contains all the real numbers
$$(S2)$$: If $$\text{Re}(a), \text{Im}(a) < 0$$, then the set $$B$$ contains all the real numbers,

Let $$a = p + qi$$ where $$p = \text{Re}(a)$$ and $$q = \text{Im}(a)$$. Let $$z = x + yi$$.

Simplify $$\text{Re}(a + \bar{z})$$.

$$a + \bar{z} = (p + qi) + (x - yi) = (p + x) + (q - y)i$$

$$\text{Re}(a + \bar{z}) = p + x$$

Simplify $$\text{Im}(\bar{a} + z)$$.

$$\bar{a} + z = (p - qi) + (x + yi) = (p + x) + (y - q)i$$

$$\text{Im}(\bar{a} + z) = y - q$$

Write the conditions for sets $$A$$ and $$B$$.

Set $$A$$: $$\text{Re}(a + \bar{z}) > \text{Im}(\bar{a} + z)$$, which gives $$p + x > y - q$$, i.e., $$x - y + (p + q) > 0$$.

Set $$B$$: $$\text{Re}(a + \bar{z}) < \text{Im}(\bar{a} + z)$$, which gives $$p + x < y - q$$, i.e., $$x - y + (p + q) < 0$$.

Check statement (S1).

Given $$\text{Re}(a) > 0$$ and $$\text{Im}(a) > 0$$, so $$p > 0$$ and $$q > 0$$.

For a real number $$z = x$$ (where $$y = 0$$), the condition for $$A$$ becomes: $$x + (p + q) > 0$$, i.e., $$x > -(p + q)$$.

Since $$p + q > 0$$, we have $$-(p + q) < 0$$. So for any real $$x \leq -(p + q)$$, the condition fails.

Therefore, $$A$$ does NOT contain all real numbers. (S1) is FALSE.

Check statement (S2).

Given $$\text{Re}(a) < 0$$ and $$\text{Im}(a) < 0$$, so $$p < 0$$ and $$q < 0$$.

For a real number $$z = x$$ (where $$y = 0$$), the condition for $$B$$ becomes: $$x + (p + q) < 0$$, i.e., $$x < -(p + q) = |p| + |q| > 0$$.

For any real $$x \geq |p| + |q|$$, the condition fails.

Therefore, $$B$$ does NOT contain all real numbers. (S2) is FALSE.

Since both (S1) and (S2) are false, the correct answer is Option D: Both are false.