(a) AC source connected to a ideal capacitor:Using KVL,
${\mathit{V}}_0\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}=\frac{\mathit{q}}{\mathit{C}}$

Or, $\mathit{I}=\frac{\mathit{d}\mathit{q}}{\mathit{d}\mathit{t}}=\mathit{C}{\mathit{V}}_0\mathit{\omega }\cos \mathit{\omega }\mathit{t}$
$\therefore \mathit{I}={\mathit{I}}_0\cos \mathit{\omega }\mathit{t}={\mathit{I}}_0\mathit{s}\mathit{i}\mathit{n}\left(\mathit{\omega }\mathit{t}+\frac{\mathit{\pi }}{2}\right)$
Average Power $=\frac{1}{\mathit{T}}\underset{0}{\overset{\mathit{T}}{\int }}\left[{\mathit{V}}_0\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}\times {\mathit{I}}_0\mathit{s}\mathit{i}\mathit{n}\left(\mathit{\omega }\mathit{t}+\frac{\mathit{\pi }}{2}\right)\right]\mathrm{dt}$
$=\frac{1}{\mathit{T}}\underset{0}{\overset{\mathit{T}}{\int }}\left[\frac{1}{2}{\mathit{V}}_0{\mathit{I}}_0\mathit{s}\mathit{i}\mathit{n}2\mathit{\omega }\mathit{t}\right]\mathrm{dt}$
$=0$
(b) Reactance of a capacitor $={\mathit{X}}_{\mathit{C}}=\frac{1}{2}\mathit{\pi }\mathit{f}\mathit{C}$
For DC, $\mathit{f}=0,{\mathit{X}}_{\mathit{C}}=\propto$, No current flow, bulb does not glow.
For $\mathrm{AC},\mathit{f}\ne 0,{\mathit{X}}_{\mathit{C}}\ne \propto$, current flows, bulb glows.
Capacitance is reduced.
For DC, $\mathit{f}=0,{\mathit{X}}_{\mathit{C}}=\propto$, No current flow, bulb does not glow.
For $\mathrm{AC}$, as $\mathrm{C}$ reduces, ${\mathrm{X}}_{\mathrm{C}}$ increases, current decreases, bulb glows less bright.