#### Answer

(a) $\frac{V_{2}}{V_{1}}=3$
(b) $\frac{E_{2}}{E_{1}}=1$

#### Work Step by Step

(a) We know that the electric potential inside the capacitor is $V=Es$, where $s$ is the distance between the point to the negative plate of the capacitor.
In the given figure , the point $1$ is at a distance of $1mm$ , thus, $s_{1}=1mm$ and the point $2$ is at a distance of $3mm$, thus, $s_{2}=1mm$.
Now, $\frac{V_{2}}{V_{1}}=\frac{s_{2}}{s_{1}}=\frac{3}{1 }$
Hence, $\frac{V_{2}}{V_{1}}=3$
(b) The electric field inside a parallel plate capacitor is constant $E=\frac{\sigma}{\epsilon_{0}}$
$E_{1}=\frac{\sigma}{\epsilon_{0}}$ and $E_{2}=\frac{\sigma}{\epsilon_{0}}$
Therefore, $ \frac{E_{2}}{E_{1}}=1$.