(a) When $q_1$ is placed, potential at $r_2=V_2=k \frac{q_1}{r_{12}}$ Potential energy when $q_2$ is placed at $r_2$ $=U_1=q_2 V_2=k q_2 \frac{q_1}{r_{12}}$ When $q_2$ is placed, potential at $r_1=V_1=k \frac{q_2}{r_{12}}$ Potential energy when $q_1$ is placed at $r_1$ $=U_2=q_1 V_1=k q_1 \frac{q_2}{r_{12}}$ Potential energy of the system $=\frac{1}{2} \left[ k q_2 \frac{q_1}{r_{12}}+k q_1 \frac{q_2}{r_{12}} \right]$ $=k \frac{q_1 q_2}{r_{12}}$ (b) An external field $E$ is applied on the system : $q_1$ and $q_2$ are two charges located at $r_1$ and $r_2$. in an external electric field (E).Work done in bringing $q_1$ from infinity to $r_1=W_1=$ $q_1 V_{r1}$Work done on $q_2$ for bringing it from infinity to $r_2$ against the external field $(E)=W_2=q_2 V_{r2}$Work done on $q_2$ against the field due to $q_1=W_{12}=\frac{q_1 q_2}{4\pi\varepsilon_0 r_{12}} (r_{12}=$ distance between $q_1$ and $q_2)$So, the potential energy of the system $=W_1+W_2+$ $W_{12}$$U=q_1 V_{r_1}+q_2 V_{r_2}+\frac{q_1 q_2}{4\pi\varepsilon_0 r_{12}}$