Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Let's assume that the circuit has negligible (zero) resistance, so the energy is conserved.
The 1200$\mu$F- capacitor (the right capacitor) will have a maximum voltage when the whole energy stored on the 300$\mu$F- capacitor is fully transferred to it.
Recalling that the energy stored in a capacitor is given by $\frac{1}{2}C(\Delta V_C)^2$.
Thus,
$$\frac{1}{2}C_{\rm left}(\Delta V_C)_{\rm left}^2=\frac{1}{2}C_{\rm right}(\Delta V_C)_{\rm right}^2$$
$$(\Delta V_C)_{\rm right}=\sqrt{\dfrac{C_{\rm left}(\Delta V_C)_{\rm left}^2}{C_{\rm right}}}$$
Plug the known;
$$(\Delta V_C)_{\rm right}=\sqrt{\dfrac{(300)(100)^2}{(1200)}}$$
$$(\Delta V_C)_{\rm right}=\color{red}{\bf 50}\;\rm V$$
$$\color{blue}{\bf [b]}$$
$\bullet$ First we need to close $\rm S_1$ at $t=0$ s.
- Now we have a $LC$ circuit, let's call it the left circuit. The charge oscillates in this circuit between the capacitor and the inductor at a periodic time of $T_{\rm left}$.
- After a time of $t_1=\frac{1}{4}T_{\rm left}$, the capacitor will be fully uncharged and the whole current is maximum through the inductor.
$\bullet$ At this instant, we need to open $\rm S_1$ and close $\rm S_2$.
- Now we have a $LC$ circuit, let's call it the right circuit. The charge oscillates in this circuit between the capacitor and the inductor at a periodic time of $T_{\rm right}$.
- After a time of $t_2=\frac{1}{4}T_{\rm left}+\frac{1}{4}T_{\rm right}$, the capacitor will be fully charged.
$\bullet$ At this instant, we need to open $\rm S_2$.
Now we need to find $t_1$ and $t_2$ where $T=2\pi/\Omega=2\pi \sqrt{LC}$.
$$t_1=\frac{1}{4}T_{\rm left}=\frac{1}{4}(2\pi \sqrt{LC_{\rm left}})$$
$$t_1= \frac{1}{2}\pi \sqrt{(5.3)(300\times 10^{-6})}=\color{red}{\bf 0.0626}\;\rm s$$
$$t_2=\frac{1}{4}T_{\rm left}+\frac{1}{4}T_{\rm right}=\frac{1}{4}(2\pi \sqrt{LC_{\rm left}})+\frac{1}{4}(2\pi \sqrt{LC_{\rm right}})$$
$$t_2= \frac{1}{2} \pi \sqrt{LC_{\rm left}} +\frac{1}{2}\pi \sqrt{LC_{\rm right}}$$
$$t_2=\frac{1}{2}\pi \sqrt{(5.3)(300\times 10^{-6})}+\frac{1}{2}\pi \sqrt{(5.3)(1200\times 10^{-6})}$$
$$t_2 =\color{red}{\bf 0.1879}\;\rm s$$
