Let $a=\alpha+i\beta,\qquad z=x+iy$ where $\alpha=\operatorname{Re}(a)$, $\beta=\operatorname{Im}(a)$ and $x,y\in\mathbb R$.

We will simplify the conditions defining $A$ and $B$.

1. Simplify set $A$

We have $\bar z=x-iy,\qquad \bar a=\alpha-i\beta.$ Then $a+\bar z=(\alpha+i\beta)+(x-iy)=(\alpha+x)+i(\beta-y),$ so $\operatorname{Re}(a+\bar z)=\alpha+x.$

Also, $\bar a+z=(\alpha-i\beta)+(x+iy)=(\alpha+x)+i(y-\beta),$ so $\operatorname{Im}(\bar a+z)=y-\beta.$

Thus the condition for $A$ is $\alpha+x>y-\beta$ which gives $x-y+\alpha+\beta>0.$ Hence $A=\{z=x+iy: x-y+\alpha+\beta>0\}.$

2. Simplify set $B$

The statement for $B$ is truncated in the prompt, but from the visible part it begins exactly the same way: $B=\{z\in\mathbb C: \operatorname{Re}(a+\bar z)(\cdots)$ The natural complementary comparison here is $\operatorname{Re}(a+\bar z)<\operatorname{Im}(\bar a+z),$ so let us evaluate that.

Then $\alpha+x<y-\beta$ which gives $x-y+\alpha+\beta<0.$ Hence $B=\{z=x+iy: x-y+\alpha+\beta<0\}.$

3. Check statement (S1)

If $\operatorname{Re}(a),\operatorname{Im}(a)>0$, then the set $A$ contains all the real numbers.

A real number $z=t\in\mathbb R$ means $x=t,\quad y=0.$ So for real $z=t$, the condition for $A$ becomes $t+\alpha+\beta>0.$

For $A$ to contain all real numbers, this inequality must hold for every $t\in\mathbb R$. But that is impossible, because for sufficiently large negative $t$, we get $t+\alpha+\beta<0.$ For example, choose $t=-(\alpha+\beta+1).$ Then $t+\alpha+\beta=-1<0.$ So this real number is not in $A$.

Therefore, (S1) is false.

4. Check statement (S2)

If $\operatorname{Re}(a),\operatorname{Im}(a)<0$, then the set $B$ contains all the real numbers.

Again let $z=t\in\mathbb R$. Then the condition for $B$ becomes $t+\alpha+\beta<0.$

For $B$ to contain all real numbers, this must hold for every real $t$. But this is impossible, because for sufficiently large positive $t$, $t+\alpha+\beta>0.$ For example, choose $t=-(\alpha+\beta)+1.$ Then $t+\alpha+\beta=1>0,$ so this real number is not in $B$.

Therefore, (S2) is false.

5. Conclusion

Both statements are false.

So the correct option is $\boxed{A}$

6. Comparison with stored answer

Stored correct answer: $A$

Our derived answer also gives $A$, so they agree.