Does the enthalpy of solution formula (LE + Hyd) change depending on the question?
Apparently your problem is related to the improper use of negative signs. Let's solve it differently.
Suppose you start from the separated ions $\ce{Ca^{2+}}$ and $\ce{2 F^-}$ in the vacuum. This state has the energy zero by definition.
Now suppose you join the ions to produce $1$ mole of solid $\ce{CaF2}$.  The process is exothermic, and the final state $\ce{CaF2(s)}$ is at the energy level $-2611 kJ/mol.$
Now suppose you are doing another transformation with the same ions. You put them directly into water, so that the ions become hydrated in solution. You obtain a solution of $\ce{CaF2}$ in water, or a solution of $\ce{Ca^{2+}(aq)}$ and $\ce{2 F^-(aq)}$. This transformation is also exothermic, and at the end $\ce{Ca^{2+}(aq)}$ + $\ce{2 F^-(aq)}$ are at the energy level $- 2662 kJ/mol$.
The final levels after these two transformations are $-2611 kJ/mol$ for the final solid, and $-2662 kJ/mol$ for the solution. These energy levels are not very different from one another.
So the passage from the solid state $\ce{CaF2}$ to the solution $\ce{Ca^{2+}(aq)}$ + $\ce{2 F^-(aq)}$ is slightly exothermic, with $\Delta H = -51 kJ/mol$