Answer
a.) $\Delta V_{max} = 50$ Volts
b.) To achieve the maximum potential difference across the $1200$ farad capacitor, we must completely discharge the $300$ farad capacitor and make sure no energy is being stored in the inductor. We can do this by first closing switch 1 for a quarter period to transfer all of the energy to the inductor and as soon as the $300$ farad capacitor is completely discharged, we reopen switch 1 and close switch 2 until the inductor has no more energy and the $1200$ farad capacitor is fully charged.
The time at which we open switch 1 and close switch 2 is $t = 0.063$ s
The time at which we open switch 2 is $t = 0.19$ s
Work Step by Step
a.) Let us first discuss how we are going to do this. We will calculate the times at which we do the following operations later in part b.)
To achieve the maximum potential difference across the $1200$ farad capacitor, we must completely discharge the $300$ farad capacitor and make sure no energy is being stored in the inductor. We can do this by first closing switch 1 for a quarter period to transfer all of the energy to the inductor and as soon as the $300$ farad capacitor is completely discharged, we reopen switch 1 and close switch 2 until the inductor has no more energy and the $1200$ farad capacitor is fully charged.
Recall that the equation for the charge on a capacitor in an $LC$ circuit is $\quad Q(t) = Q_0\cos(\omega t)$
and that the current in the circuit is $\quad I(t) = Q_0\omega\cos(\omega t) = I_{max}\cos(\omega t)$
where the angular frequency is $\quad\omega = \displaystyle \frac{1}{\sqrt{LC}}$
In both processes of discharging the $300$ farad capacitor and fully charging the $1200$ farad capacitor, the maximum current in both $LC$ circuits will be the same. Note though, that the $300$ farad capacitor-inductor LC circuit will have its maximum current exactly when the $300$ farad capacitor is fully discharged and at the exact same time that we reopen switch 1 and close switch 2. The $1200$ farad capacitor-inductor circuit will have that same maximum current at the exact instant that switches 2 is closed (so basically the same time). Nonetheless, this means (Note that both $Q_{300}$ and $Q_{1200}$ are the maximum charges for their respective capacitors)
$I_{max, 300LC} = I_{max, 1200LC}$
$Q_{300}\omega_{300} = Q_{1200}\omega_{1200}$
$\displaystyle \frac{Q_{300}}{\sqrt{LC_{300}}} = \frac{Q_{1200}}{\sqrt{LC_{1200}}}$
$\displaystyle Q_{300}\sqrt{\frac{C_{1200}}{C_{300}}} = Q_{1200}$
$Q_{1200} = (300\cdot10^{-6})(100)\displaystyle \sqrt{\frac{1200\cdot10^{-6}}{300\cdot10^{-6}}} = (300\cdot10^{-4})\sqrt{4} = 0.03(2) = 0.06$ C
Now, we know that when the maximum voltage across the $1200$ farad capacitor will happen when it has its maximum charge (note the maximum value for $\cos_{max}(\omega t) = 1$ and that this will happen when the $1200$ capacitor has its maximum charge) :
$Q(t) = Q_0\cos(\omega t) = C\Delta V\cos(\omega t)$
$Q_{max} = C\Delta V_{max} (1)$
$\Delta V_{max} = \displaystyle \frac{Q_{1200}}{C_{1200}} = \frac{0.06}{1200\cdot10^{-6}} = 50$ Volts
OR we can just compare the energies in the capacitors (much easier):
$\displaystyle \frac{1}{2}C_1\Delta V_1^2 = \frac{1}{2}C_2\Delta V_2^2$
$\Delta V_1 \displaystyle \sqrt{\frac{C_1}{C_2}} = \Delta V_2 = 50$ Volts
b.) The time at which we open switch 1 and close switch 2 is when the $300$ farad capacitor has no more charge (fully discharged):
$Q_{300}(t) = 0 = Q_{300}\cos(\omega_{300} t)$
$\cos^{-1}(0) = \displaystyle \frac{t}{\sqrt{LC_{300}}}$
$\displaystyle \frac{\pi}{2} = \frac{t}{\sqrt{LC_{300}}}$
$\displaystyle \frac{\pi}{2}\sqrt{LC_{300}} = \frac{\pi}{2}\sqrt{(5.3)(300\cdot10^{-6})} \approx 0.063$ seconds
The time it takes after the above time to then open switch 2 after the $1200$ farad capacitor is fully charged is:
$\displaystyle Q_{1200}(t) = Q_{1200}\cos(\omega_{1200} t - \frac{\pi}{2}) = Q_{1200}$
You may ask, why is there a $\displaystyle \frac{\pi}{2}$ in your equation for charge? That is because we must realize that the charge on the $1200$ farad capacitor is not at its max charge at time $t = 0$. By subtracting the $\displaystyle \frac{\pi}{2}$ in the equation, we see that we are phase-shifting the trigonometric function so that at time $t = 0$ that $Q_{1200} = 0$ like it should be (plug in $0$ for $t$ with the $\displaystyle \frac{\pi}{2}$ in the equation to see that the charge is then 0).
$1 = \cos(\omega_{1200} t - \displaystyle \frac{\pi}{2})$
$\cos^{-1}(1) = \displaystyle \frac{t}{\sqrt{LC_{1200}}} - \frac{\pi}{2}$
$0 = \displaystyle \frac{t}{\sqrt{LC_{1200}}} - \frac{\pi}{2}$
$\displaystyle \frac{t}{\sqrt{LC_{1200}}} = \frac{\pi}{2}$
$t = \displaystyle \frac{\pi}{2}\sqrt{LC_{1200}} = \frac{\pi}{2}\sqrt{(5.3)(1200\cdot10^{-6})} \approx 0.13$ s
But realize that this is just the time after we close switch 2, so the total time after we first close switch 1 to when we finally reopen switch 2 is
$0.062635 + 0.125270 = 0.187906$ seconds
