Determining enthalpy and internal energy of formation for potassium superoxide
Now, when we say that the temperature rose by $\pu{2.62K}$, this is the temperature of the calorimeter and the water together. We are assuming that they have been given time to reach thermal equilibrium (and generally water and copper calorimeters take relatively less time to reach equilibrium, it's pretty safe to neglect the time). Calorimeters can be filled with varied amounts of water, so the calorimeter's constant is for just the calorimeter, not the water in it.

Note that if we were to assume that the specific heat of potassium/oxide were not negligible, we would have to know the specific heats of the two, as well as the temperature at the start of the reaction and the temperature at which they want $\Delta H^o$ to be calculated at. "Standard state" doesn't fix temperature, though when you say $\Delta H^o$, you generally mean $\Delta H^o_{\pu{298K}}$. Using these, you can form a reaction pathand calculate the answer (I can elaborate on this if you want, it's pretty interesting)

With these assumptions, one gets the heat released to the order of 10 kJ, and the energy is to the order of 100 kJ/mol. Not too extravagant, remember that a joule is a rather tiny unit of energy. And formation enthalpies/energies are generally to the order of 100 or 1000 kJ.

Finding the enthalpy change now is pretty easy. Since $\Delta H=\Delta U+ \Delta(PV)=\Delta U+\Delta(nRT)$. Using this and looking at the overall reaction, it shouldn't be too hard :)