Answer
(a) $Q = 115.2 \mathrm{~\mu C}$
(b) $V_1 = 38.4 \,\text{V}$ and $V_2 = 23.04\,\text{V}$
Work Step by Step
(a) $C_1$ and $C_2$ are in series, which means they have the same charge and their equivalent capacitance is calculated by
$$\dfrac{1}{C_{12}} = \dfrac{1}{C_{1} } + \dfrac{1}{C_{2} } = \dfrac{1}{3\mathrm{~\mu F} } + \dfrac{1}{5 \mathrm{~\mu F} } = 0.533 $$
Take the reciprocal
$$C_{12} = 1.8 \mathrm{~\mu F} $$
Therefore, the charge on each plate is given by
$$Q = C_{12} V = (1.8 \mathrm{~\mu F} ) (64 \,\text{V}) = \boxed{115.2 \mathrm{~\mu C}}$$
(b) The potential in the series connection is different for each capacitor.
For $C_1$ the potential is given by
$$V_1 = \dfrac{Q}{C_1} = \dfrac{115.2 \mathrm{~\mu C}}{3\mathrm{~\mu F}} = 38.4 \,\text{V}$$
For $C_2$ the potential is given by
$$V_2 = \dfrac{Q}{C_2} = \dfrac{115.2 \mathrm{~\mu C}}{5\mathrm{~\mu F}} = 23.04 \,\text{V}$$