(i) The extension of the spring is $(d_0-d_1)$ so the additional upward force exerted by the spring at equilibrium (in excess of the force $mg$ required to support the uncharged plate) is $K(d_0-d_1)$.

At equilibrium the electric potential between the plates is $V$ and the separation is $d_1$. The electric field is uniformly $E=V/d_1$. The charge on each plate is $Q=CV=\epsilon A V/d_1$. Each plate contributes $\frac12 E$ to the electric field between the plates. Each plate only experiences the electric field due to the other plate. So the force of attraction pulling the top plate down is $\frac12 QE=\epsilon AV^2/2d_1^2$.

Equating these two forces gives

$K=\epsilon AV^2/2(d_0-d_1)d_1^2$.

(ii) Rearrange the force-balance equation to get

$V^2=B(d_0d_1^2-d_1^3)$

where $B$ is a constant. Then differentiate to find the maximum value of $V$ as a function of $d_1$.

(iii) Suppose $d_1$ increases by a small amount $x$. Then the upward force on the upper plate from the spring becomes $K(d_0-(d_1+x))$ and the downward force from the fixed lower plate becomes $\epsilon AV^2/2(d_1+x)^2 \approx (\epsilon AV^2/2d_1^2)(1-2\frac{x}{d_1})$ using the binomial expansion and assuming that $x \ll d_1$. The equation of motion of the upper plate is then

$m\ddot x = K(d_0-d_1-x)-\frac{\epsilon AV^2}{2d_1^2}(1-2\frac{x}{d_1})$

$m\ddot x +(K-\frac{\epsilon AV^2}{d_1^3})x=K(d_0-d_1)-\frac{\epsilon AV^2}{2d_1^2}=0$

because the forces balance at equilibrium separation $d_1$. This equation is of the form $\ddot x + \omega^2x = 0$ so the motion is (approx) SHM with angular frequency $\omega$, where

$\omega^2 =\frac{K}{m}-\frac{\epsilon AV^2}{md_1^3}$.