Answer
(a) $Q_1 = 60 \mathrm{~\mu C}$ and $Q_3 = 90 \mathrm{~\mu C}$
(b) $V_{ab} = 28 \,\text{V}$
Work Step by Step
(a) In parallel, both $C_1$ and $C_2$ have the same potential. So, we could get the next using the potential as $V = \dfrac{Q}{C}$
\begin{gather*}
\dfrac{Q_1}{C_1} = \dfrac{Q_2}{C_2} \\
Q_1 = \dfrac{Q_2}{C_2}\,C_1 \\
Q_1 = \dfrac{30 \mathrm{~\mu C}}{3 \mathrm{~\mu F}}\,6 \mathrm{~\mu F}\\
Q_1 = 60 \mathrm{~\mu C}
\end{gather*}
The total charge of the combination $C_{12}$ is given by
$$Q_{12} = Q_1 + Q_2 = 60 \mathrm{~\mu C} + 30 \mathrm{~\mu C} = 90 \mathrm{~\mu C}$$
$C_3$ and the combination $C_{12}$ are in series, where, in series both elements have the same charge, therefore, the charge of $C_3$ is
$$Q_3 = \boxed{90 \mathrm{~\mu C}}$$
(b) The combination $C_{12}$ has a capacitance
$$C_{12} = C_1 + C_2 = 6 \mathrm{~\mu F} + 3 \mathrm{~\mu F} = 9 \mathrm{~\mu F}$$
$C_{12}$ and $C_3$ are in series, which means they have the same charge and their equivalent capacitance is calculated by
$$\dfrac{1}{C_{eq}} = \dfrac{1}{C_{12} } + \dfrac{1}{C_{3} } = \dfrac{1}{9\mathrm{~\mu F} } + \dfrac{1}{5 \mathrm{~\mu F} } = 0.311$$
Take the reciprocal
$$C_{eq} = 3.21 \mathrm{~\mu F} $$
Now it is easy to find the potential between $a$ and $b$ as next
$$V_{ab} = \dfrac{Q}{C_{eq}} = \dfrac{90 \mathrm{~\mu C}}{3.21 \mathrm{~\mu F} }= \boxed{28 \,\text{V}}$$
