Does enthalpy of dissolution change with temperature?
I want to know whether an increase in temperature affects the enthalphy of dissolution? 


It certainly can. You can treat the temperature dependence of dissolution with a formalism analogous to that applied to chemical reactions. The key parameter is the difference in heat capacity of the solid and dissolved solute, since the enthalpy of solubilization can be written as 

$$\Delta H(T) = \Delta H(T_\mathrm{ref}) + \int_{T_\mathrm{ref}}^{T} \Delta C_p\,\mathrm{d}T$$

where $\Delta C_p = C_{p,\mathrm{solute}} - C_{p,\mathrm{s}}$. The $\Delta C_p$ term accounts for the different temperature sensitivities of the enthalpy (which for condensed phases approximates the energy) of solid and solute. If the heat capacity of the solute is greater than that of the solid, the enthalpy of dissolution will increase (become more positive) with an increase in $T.$

Note that if we associate a solubility constant $K_\mathrm{sp}$ with the dissolution process

$$\ce{A(solid) &lt;=&gt; A(solution)}$$

then we can write

$$\frac{\mathrm{d}\log K_\mathrm{sp}}{\mathrm{d}T}=\frac{\Delta H^\circ}{RT^2}$$

as suggested in a comment. Here both $K_\mathrm{sp}$ and $\Delta H^\circ$ are functions of $T.$