Answer
a) $\rm -10\;V/m,-10\;V/m$
b) $\rm -5\;V/m,-20\;V/m$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We also know that the electric field is given by
$$E_x=-\dfrac{dV}{dx}$$
So at point 1, from the given graph,
$$\vec E_{1}=-\dfrac{dV}{dx}\;\hat i=-\dfrac{10-0}{1-0}\;\hat i$$
$$\vec E_{1}=-(\color{red}{\bf10}\;{\rm V/m})\;\hat i$$
At point 2, from the given graph,
$$\vec E_{2}=-\dfrac{dV}{dx}\;\hat i=-\dfrac{40-30}{4-3}\;\hat i$$
$$\vec E_{2}=-(\color{red}{\bf10}\;{\rm V/m})\;\hat i$$
$$\color{blue}{\bf [b]}$$
At point 1, from the given graph,
$$\vec E_{1}=-\dfrac{dV}{dx}\;\hat i=-\dfrac{10-0}{2-0}\;\hat i$$
$$\vec E_{1}=-(\color{red}{\bf 5}\;{\rm V/m})\;\hat i$$
At point 2, from the given graph,
$$\vec E_{2}=-\dfrac{dV}{dx}\;\hat i=-\dfrac{40-20}{4-3}\;\hat i$$
$$\vec E_{2}=-(\color{red}{\bf20}\;{\rm V/m})\;\hat i$$
From all the above, it is obvious that the 4 electric fields at the 4 points are pointing left.
