Determination of ${\mathit{C}}_1$ and ${\mathit{C}}_2$
Determination of Charge on each capacitor in parallel combination
Energy stored in a capacitor
$\mathit{E}=\frac{1}{2}\mathit{C}{\mathit{V}}^2$
In series combination
$0.045=\frac{1}{2}\frac{{\mathit{C}}_1{\mathit{C}}_2}{{\mathit{C}}_1+{\mathit{C}}_2}{\left(100\right)}^2$
$\Rightarrow \frac{{\mathit{C}}_1{\mathit{C}}_2}{{\mathit{C}}_1+{\mathit{C}}_2}=0.09\times {10}^{-4}$ ......(i)
In Parallel combination
$0.25=\frac{1}{2}\left({\mathit{C}}_1+{\mathit{C}}_2\right){\left(100\right)}^2$
$\Rightarrow {\mathit{C}}_1+{\mathit{C}}_2=0.5\times {10}^{-4}$ .......(ii)
On simplifying (i) & (ii)
${\mathit{C}}_1{\mathit{C}}_2=0.045\times {10}^{-8}$
${\left({\mathit{C}}_1-{\mathit{C}}_2\right)}^2={\left({\mathit{C}}_1+{\mathit{C}}_2\right)}^2-4{\mathit{C}}_1{\mathit{C}}_2$
$={\left(0.5\times {10}^{-4}\right)}^2-4\times 0.045\times {10}^{-8}$
$=0.25\times {10}^{-8}-0.180\times {10}^{-8}$
${\left({\mathit{C}}_1-{\mathit{C}}_2\right)}^2=0.07\times {10}^{-8}$
$\left({\mathit{C}}_1-{\mathit{C}}_2\right)=2.6\times {10}^{-5}$
$=0.26\times {10}^{-4}$ .......(iii)
From (ii) and (iii) we have
$\Rightarrow {\mathit{C}}_1=0.38\times {10}^{-4}\mathrm{F}$

Charges on capacitor ${\mathit{C}}_1$ and ${\mathit{C}}_2$ in Parallel combination
${\mathit{Q}}_1={\mathit{C}}_1\mathit{V}=\left(0.38\times {10}^{-4}\times 100\right)$
$=0.38\times {10}^{-2}\mathrm{C}$
${\mathit{Q}}_2={\mathit{C}}_2\mathit{V}=\left(0.12\times {10}^{-4}\times 100\right)$
$=0.12\times {10}^{-2}\mathrm{C}$
[Note: If the student writes the relations/ equations
$\mathit{E}=\frac{1}{2}\mathit{C}{\mathit{V}}^2$
and $0.045=\frac{1}{2}\left(\frac{{\mathit{C}}_1{\mathit{C}}_2}{{\mathit{C}}_1+{\mathit{C}}_2}\right){\left(100\right)}^2$
$0.25=\frac{1}{2}\left({\mathit{C}}_1+{\mathit{C}}_2\right){\left(100\right)}^2$
But is unable to calculate ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$, award him/her full 2 marks.
Also if the student just writes

Award him/her one mark for this part of the question.]