given: Cu2O(s) + 1/2 O2(g) 2CuO(s) DH° = -144 kJ?

Question

Given:

Cu2O(s) + 1/2 O2(g) 2CuO(s) DH° = -144 kJ

Cu2O(s) Cu(s) + CuO(s) DH° = +11 kJ

Calculate the standard enthalpy of formation of CuO(s).

Answer 1

The --> ("reacts to yield") in the chemical reactions seem to be omitted from your question.
The reactions should read:
Cu2O(s) + 1/2 O2(g) --> 2CuO(s) DH° = -144 kJ
Cu2O(s) --> Cu(s) + CuO(s) DH° = +11 kJ

By flipping the direction of the top equation we get,
2CuO(s) --> Cu2O(s) + 1/2 O2(g) DH° = +144 kJ
Now combine this with the second equation,
2CuO(s) + Cu2O(s) --> Cu(s) + CuO(s) + Cu2O(s) + 1/2 O2(g) DH° = +155 kJ
Now cancel out anything which is on both sides of the reaction,
and we are left with,
CuO(s) --> Cu(s) + 1/2 O2(g) DH° = +155 kJ
The standard enthalpy of formation for Copper metal and Oxygen gas are both zero since they are in their elemental states in this reaction so that leaves up with a value of the standard enthalpy of formation of -155 kJ/mol of CuO since CuO is on the left side of the reaction we have to flip the sign on the delta H value.

Final answer should be H°f of CuO = -155 kJ/mole.

*To check this answer I looked up the actual value in a reference table I have to find the value of -157.3 kJ/mol...not too bad considering the question probably gave up rounded off values to use in order to make the math simpler.