Can we dissolve salts with higher lattice enthalpy by providing heat?
Heat itself it just indirect factor. It is temperature what determines the shift of solubility products. If the dissolution is endothermic process, the salt solubility product would increase with higher temperature.
But if the solubility product is still small even at high temperature, like e.g. $\pu{e-16}$ instead of $\pu{e-20}$, then only a tiny extra salt amount would dissolve when heated.
The effective result depends on the particular case. The typical cases where significant solubility increase is observed are  potassium salts  $\ce{KNO3, KClO3, KClO4}$. But they have limited solubility even at low temperature - see the solubility table.
Generally, the dependence of an equlibrium constant of a reaction, including dissolving/precipitation process, is described by van't Hoff equation

$$\frac{\mathrm{d}}{\mathrm{d}T} \ln K_\mathrm{eq} = \frac{\Delta H^\ominus}{RT^2}$$

For solubility products, being equlibrium constants, is valid the following reaction:
$$\Delta G^{\circ}_\mathrm{dis} = \Delta H^{\circ}_\mathrm{dis}  - T \Delta S^{\circ}_\mathrm{dis} = RT \ln K_\mathrm{sp}$$