Answer
a) $0\;\rm N\cdot m^2/C^2$
b) $2\pi R^2E$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric flux is given by
$$ \Phi=\vec E\cdot \vec A =EA\cos\theta$$
where $A$ here consists of 3 parts, the two circular bases of the cylinder plus the side walls which is a rectangle.
Thus,
$$ \Phi=EA_{\rm left }\cos0^\circ+EA_{\rm right}\cos180^\circ +EA_{\rm side}\cos90^\circ$$
$$ \Phi=E(\pi R^2)(1)+E(\pi R^2) (-1) +E(2\pi R L)(0)$$
We know that $E$ is uniform throughout the whole cylinder.
$$ \Phi=\color{red}{\bf 0}\;\rm N\cdot m^2/C^2$$
$$\color{blue}{\bf [b]}$$
By the same approach,
$$ \Phi=EA_{\rm left }\cos0^\circ+EA_{\rm right}\cos0^\circ +EA_{\rm side}\cos90^\circ$$
$$ \Phi=E(\pi R^2)(1)+E(\pi R^2) (1) +E(2\pi R L)(0)$$
$$ \boxed{\Phi=2\pi R^2E}$$
