Two concentric shells $A and B$ have radii $R and 2 R,$ charges $q_A and q_B$and potentials $2 V$and $\frac{3}{2}$V respectively. Now, the shell $B$ is earthed and the net charges on them become $q_A^{\text{'}}$and $q_B^{\text{'}}.$ Then

$\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_A}{R}+\frac{q_B}{2R}\right)=2V$

$\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_A}{2R}+\frac{q_B}{2R}\right)=\frac{3}{2}V$

Solving equations (1) and (2), we get

$\frac{q_A}{q_B}=\frac{1}{2}$

When B is earthed, potential of B becomes zero. So, $q_B^{\text{'}}=-q_A^{\text{'}}=-q_A$charge on A remains same 

$\to \frac{q_A^{\text{'}}}{q_B^{\text{'}}}=-1$

Also after earthing,

$V_A-V_B=\frac{q_A}{4\pi {\epsilon }_0}\left(\frac{1}{R}-\frac{1}{2R}\right)=\frac{q_A}{8\pi {\epsilon }_{0}R}$

Substituting $q_B= 2q_A$ in equation (1), we get

$$\begin{gathered} \frac{1}{4\pi {\epsilon }_0}\frac{q_A}{2R}=\frac{V}{2} \\ \to V_A-V_B=\frac{V}{2} \end{gathered}$$

Since B is earthed, so $V_B=0$ and hence $V_A=\frac{V}{2}$

Hence, (1) and (4) are correct.