Answer
(a) ${C_{eq}}' = 19.3 \mathrm{~n F} $
(b) $ Q = 482.5 \mathrm{~n C}$
(c) $Q = 162.5 \mathrm{~n C}$
(d) $V = 25 \,\text{V}$
Work Step by Step
a) For the capacitor in series, they have an equivalent capacitance calculated by
$$\dfrac{1}{C_{eq}} = \dfrac{1}{18\mathrm{~ nF} } + \dfrac{1}{30 \mathrm{~nF} } +\dfrac{1}{10 \mathrm{~nF} } = 0.188 $$
Take the reciprocal
$$C_{eq} = 5.30 \mathrm{~n F} $$
For the capacitor in parallel, they have an equivalent capacitance calculated by
$${C_{eq}}' = C_{eq} + 7.5 \mathrm{~nF} + 6.5 \mathrm{~n F} = \boxed{19.3 \mathrm{~n F} }$$
(b) The charges of the system are given by
$$Q = {C_{eq}}' V_{ab}$$
Where $V_{ab}$ is the potential between the two terminals. Let us substitute the values $C$ and $V_{ab}$ to get the charge on each plate
$$Q = {C_{eq}}' V_{ab} = (19.3 \mathrm{~n F})(25 \,\text{V}) = \boxed{482.5 \mathrm{~n C}}$$
(c) $$Q = CV= (6.5 \mathrm{~n F})(25 \,\text{V}) = \boxed{162.5 \mathrm{~n C}}$$
(d) 7.5 nF has the same potential of the system as it in parallel with the remaining elements
$$\boxed{V = 25 \,\text{V}}$$
