Answer
See the detailed answer below.
Work Step by Step
$\bullet$ At point 1:
The direction of the electric field of the left sheet is toward the sheet itself which means toward the right while the direction of the electric field of the right sheet is away from the sheet which means toward the left.
Thus, the net electric field at point 1 is given by
$$E_{net}=E_{left}-E_{right}$$
where $E_{sheet}=\eta/2\epsilon_0$, so
$$E_{net}=\dfrac{ \eta_{left}}{2\epsilon_0}\;\hat i-\dfrac{ \eta_{right}}{2\epsilon_0}\;\hat i$$
$$E_{net}=\dfrac{|-\eta_0|}{2\epsilon_0}\;\hat i-\dfrac{|3\eta_0|}{2\epsilon_0}\;\hat i$$
$$\boxed{E_{net}= -\dfrac{ \eta_0 }{ \epsilon_0}\;\hat i}$$
$\bullet\bullet$ At point 2:
The direction of the electric field of the left sheet is toward the sheet itself which means toward the left while the direction of the electric field of the right sheet is away from the sheet which means toward the left as well.
Thus, the net electric field at point 1 is given by
$$E_{net}=-(E_{left}+E_{right})$$
$$E_{net}=-\left[\dfrac{ \eta_{left}}{2\epsilon_0}\;\hat i+\dfrac{ \eta_{right}}{2\epsilon_0}\;\hat i \right]$$
$$\boxed{E_{net}= -\dfrac{ 2\eta_0 }{ \epsilon_0}\;\hat i}$$
$\bullet\bullet\bullet$ At point 3:
The direction of the electric field of the left sheet is toward the sheet itself which means toward the left while the direction of the electric field of the right sheet is away from the sheet which means toward the right.
Thus, the net electric field at point 1 is given by
$$E_{net}=E_{right}-E_{left}$$
where $E_{sheet}=\eta/2\epsilon_0$, so
$$E_{net}=\dfrac{ \eta_{right}}{2\epsilon_0}\;\hat i-\dfrac{ \eta_{left}}{2\epsilon_0}\;\hat i $$
$$E_{net}=\dfrac{|3\eta_0|}{2\epsilon_0}\;\hat i-\dfrac{|-\eta_0|}{2\epsilon_0}\;\hat i$$
$$\boxed{E_{net}= \dfrac{ \eta_0 }{ \epsilon_0}\;\hat i}$$
