(a)
When ${\mathrm{q}}_1$ is placed, potential at ${\mathit{r}}_2={\mathit{V}}_2=\mathit{k}\frac{{\mathit{q}}_1}{{\mathit{r}}_{12}}$
Potential energy when ${\mathit{q}}_2$ is placed at ${\mathit{r}}_2$
$={\mathit{U}}_1={\mathit{q}}_2{\mathit{V}}_2=\mathit{k}{\mathit{q}}_2\frac{{\mathit{q}}_1}{{\mathit{r}}_{12}}$
When ${\mathit{q}}_2$ is placed, potential at ${\mathit{r}}_1={\mathit{V}}_1=\mathit{k}\frac{{\mathit{q}}_2}{{\mathit{r}}_{12}}$
Potential energy when ${\mathit{q}}_1$ is placed at ${\mathit{r}}_1$
$={\mathrm{U}}_2={\mathit{q}}_1{\mathrm{V}}_1=\mathit{k}{\mathit{q}}_1\frac{{\mathit{q}}_2}{{\mathit{r}}_{12}}$
Potential energy of the system
$=1∕2\left[\mathit{k}{\mathit{q}}_2\frac{{\mathit{q}}_1}{{\mathit{r}}_{12}}+\mathit{k}{\mathit{q}}_1\frac{{\mathit{q}}_2}{{\mathit{r}}_{12}}\right]$ $=\mathit{k}\frac{{\mathit{q}}_1{\mathit{q}}_2}{{\mathit{r}}_{12}}$
(b) An external field $\mathrm{E}$ is applied on the system : ${\mathit{q}}_1$ and ${\mathit{q}}_2$ are two charges located at ${\mathit{r}}_1$ and ${\mathit{r}}_2$. in an external electric field (E).
Work done in bringing ${\mathit{q}}_1$ from infinity to ${\mathit{r}}_1={\mathit{W}}_1=$ ${\mathit{q}}_1{\mathrm{V}}_{\mathrm{r}1}$
Work done on ${\mathit{q}}_2$ for bringing it from infinity to ${\mathit{r}}_2$ against the external field $\left(\mathrm{E}\right)={\mathrm{W}}_2={\mathit{q}}_2{\mathrm{V}}_{\mathrm{r}2}$
Work done on ${\mathit{q}}_2$ against the field due to ${\mathit{q}}_1={\mathit{W}}_{12}=\frac{{\mathit{q}}_1{\mathit{q}}_2}{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}_{12}}({\mathit{r}}_{12}=$ distance between ${\mathit{q}}_1$ and ${\mathit{q}}_2)$
So, the potential energy of the system $={\mathit{W}}_1+{\mathit{W}}_2+$ ${\mathrm{W}}_{12}$
$\mathit{U}={\mathit{q}}_1{\mathit{V}}_{{\mathit{r}}_1}+{\mathit{q}}_2{\mathit{V}}_{{\mathit{r}}_2}+\frac{{\mathit{q}}_1{\mathit{q}}_2}{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}_{12}}$