Answer
(a) $\frac{T}{4}$
(b) $\frac{T}{2}$
Work Step by Step
(a) We know that the current passing though the capacitor is given as
$i=\frac{dq}{dt}$
$\implies i=\frac{d}{dt}(q_m cos\omega t)$
$\implies i=-\omega q_m sin\omega t$
We also know that for the maximum value of $i$, the value of $sin\omega t$ is equal to $1$.
$\implies sin\omega t=1$
$\implies sin\omega t=sin \frac{\pi}{2}$
Thus $\omega t=\frac{\pi}{2}$
$\implies t=\frac{\pi}{2\omega}$
$\implies t=\frac{\pi}{2(\frac{2\pi}{T})}$
$t=\frac{t}{4}$
We conclude that after $\frac{T}{4}$ seconds the current in the circuit will be maximum.
(b) We know that
$t=\frac{n\pi}{\frac{2\pi}{T}}$
$\implies t=\frac{nT}{2}$
For $n=0$ we have $t=0$
For $n=1$ then we have $t=\frac{(1)T}{2}$
Thus, after $\frac{T}{2}$ seconds, the energy stored in the electric field is maximum.
