The following operations can be carried out on a capacitor: X –...

Let the initial capacitance be $C$ and the dielectric constant of the slab be $K > 1$.

For option 1:
Sequence WXY: Introducing the dielectric (W) makes the capacitance $KC$. Attaching the battery (X) charges it to a potential $\mathcal{E}$, giving stored energy $U_1 = \dfrac{1}{2}KC\mathcal{E}^2$. Disconnecting the battery (Y) keeps the charge and energy constant. Thus, $U_1 = \dfrac{1}{2}KC\mathcal{E}^2$.
Sequence XYW: Attaching the battery (X) charges the capacitor to $Q = C\mathcal{E}$. Disconnecting the battery (Y) keeps the charge constant at $C\mathcal{E}$. Introducing the dielectric (W) increases the capacitance to $KC$. The new energy is $U_2 = \dfrac{Q^2}{2(KC)} = \dfrac{(C\mathcal{E})^2}{2KC} = \dfrac{C\mathcal{E}^2}{2K}$.
Since $K > 1$, $U_1 > U_2$. Option 1 is correct.

For option 2:
Sequence XW: The battery is attached (X), so the potential difference is $\mathcal{E}$. Introducing the dielectric (W) while the battery is still connected keeps the potential difference at $\mathcal{E}$. The electric field is $E_1 = \dfrac{\mathcal{E}}{d}$.
Sequence WX: The dielectric is introduced (W), then the battery is attached (X). The potential difference is again $\mathcal{E}$, and the electric field is $E_2 = \dfrac{\mathcal{E}}{d}$.
Thus, $E_1 = E_2$. Option 2 is correct.

For option 3:
Sequence XWY: The battery is attached (X) and the dielectric is introduced (W). The capacitance becomes $KC$ and the potential difference is $\mathcal{E}$. The charge is $Q_1 = KC\mathcal{E}$. Disconnecting the battery (Y) keeps this charge constant.
Sequence XYW: The battery is attached (X), giving charge $Q_2 = C\mathcal{E}$. Disconnecting the battery (Y) and then introducing the dielectric (W) keeps the charge constant at $Q_2 = C\mathcal{E}$.
Since $K > 1$, $Q_1 > Q_2$. Option 3 is correct.

For option 4:
Sequence XYZ: Attaching the battery (X) gives an initial charge $Q_i = C\mathcal{E}$ and stored energy $U_i = \dfrac{1}{2}C\mathcal{E}^2$. Disconnecting (Y) keeps these constant. Reconnecting with reversed polarity (Z) changes the charge on the plates to $Q_f = -C\mathcal{E}$. The final energy is $U_f = \dfrac{1}{2}C(-\mathcal{E})^2 = \dfrac{1}{2}C\mathcal{E}^2$. The stored energy stays the same.
However, a charge of $\Delta Q = 2C\mathcal{E}$ flows through the battery to reverse the polarity. The work done by the battery is $W = (2C\mathcal{E})\mathcal{E} = 2C\mathcal{E}^2$.
The heat produced is $H = W - (U_f - U_i) = 2C\mathcal{E}^2 - 0 = 2C\mathcal{E}^2$.
Since thermal energy is produced, option 4 is incorrect.