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

Let $a=x_1+i{y}_1, \overline{a}=x_1-i{y}_1$

And $z=x_2+i{y}_2, \overline{z}=x_2-i{y}_2$      

Now,

$a+\overline{z}=\left(x_1+x_2\right)+i\left(y_1-y_2\right)$

and $\overline{a}+z=\left(x_1+x_2\right)+i\left(y_2-y_1\right)$

Now according to the question,

$Re(a+\overline{z})=x_1+x_2$ and

$Im(\overline{a}+z)=y_2-y_1$

$A=\left\{z:x_1+x_2>y_2-y_1\right\}=\left\{z:x_1+y_1+x_2>y_2\right\}$
  $B=\left\{z:x_1+x_2<y_2-y_1\right\}=\left\{z:x_1+y_1+x_2<y_2\right\}$

If  $y_2=0$ and $x_1,y_1>0$ then

$A=\{z:x_2>-\left(x_1+y_1\right)\}$

A covers a part of negative real axis and therefore, does not contain whole real axis

Similarly if $y_2=0$, and $x_1,y_1<0$, then 
$B=\{z:x_2<-\left(x_1+y_1\right)\}$
$\therefore$ $B$ covers part of positive real axis and therefore does not cover whole real axis.

Hence, both are false.