From the given information

$\Delta \mathrm{H}$ is negative so it means dissolution of gas $\mathrm{HCl}(\mathrm{g})$ is exothermic.

$$\mathrm{HCl}(\mathrm{~g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}\quad\text{..... (1)}$$

$$\begin{aligned} & \Delta \mathrm{H}_1=-69.01 \frac{\mathrm{~kJ}}{\mathrm{~mol}} \\ & \mathrm{HCl}(\mathrm{~g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}\quad\text{..... (2)} \end{aligned}$$

$$\Delta \mathrm{H}_2=-72.79 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$$

Hence heat of solution depends upon amount of solvent

By equation ............. (2) $-$ equation ........... (1)

$$\mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}+30 \mathrm{H}_2 \mathrm{O}(\ell) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}$$

So Heat of dilution $=-72.79-(-69.01)$

$$=-3.78 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$$

Hence option (3) is incorrect.

For heat of formation reactant should be in elemental form hence option (4) is incorrect