The sign of enthalpy of formation of magnesium oxide
I guess you've been answered in the comments already, but anyway. You are correct. In thermodynamics, enthalpy is the energy supplied to the system at constant pressure. If a reaction is accompanied by the release of heat, i.e. if it is exothermic, its $\Delta H$ is negative. If $x$ kJ is released into the calorimeter upon dissolving 1 mol of substance, resulting in the rise of calorimeter temperature, then $\Delta H_{diss} = -x$ kJ/mol.

As for the heat, depends on the type of the calorimeter, i.e. on the direction in which the heat flows. For conventional "coffee cup" calorimeter, where $Q=mC\Delta T$, $Q$ is the energy released, so $Q = -\Delta H$. For some power-compensated calorimeter, where you vary the heater power so as to maintain the temperature, $Q$ would be equal to the energy supplied. Since you have to supply excess energy ($Q&gt;0$) to offset the endothermic reaction ($\Delta H&gt;0$), $Q = \Delta H$.

I've heard that a long time ago, at least in some countries, there were two systems of signs for enthalpies, so-called "thermodynamic" and "thermochemical," with reversed signs in the latter. However, I haven't seen "thermochemical" one in real life, and it seems to be long since abandoned. It is not even mentioned in modern textbooks on thermodynamics. So only the thermodynamic sign system, where $\Delta H&gt;0$ for endothermic processes, remains.