stuck on this hess's law question?

You need "3FeO(s)" on the left. The only given equation that mentions FeO(s) is the third one. So multiply the third given equation by 3, then write it backwards:
3 FeO(s) + 3 CO(g) → 3 Fe(s) + 3 CO2(g), ΔH° = -33.0 kJ
You also need "Fe3O4(s)" on the right. The only given equation that mentions "Fe3O4(s)" is the first one. So divide the first given equation by 2, then write it backwards:
3/2 Fe2O3(s) + 1/2 CO(g) → Fe3O4(s) + 1/2 CO2(g), ΔH° = -19.5 kJ
Add the last two equations here:
3 FeO(s) + 3 CO(g) + 3/2 Fe2O3(s) + 1/2 CO(g) →
3 Fe(s) + 3 CO2(g) + Fe3O4(s) + 1/2 CO2(g), ΔH° = -33.0 kJ -19.5 kJ
Combine like amounts on the same side of the arrow, and do the arithmetic for ΔH°:
3 FeO(s) + 7/2 CO(g) + 3/2 Fe2O3(s) → 3 Fe(s) + 7/2 CO2(g) + Fe3O4(s), ΔH° = -52.5 kJ
Now you need to get rid of "3/2 Fe2O3(s)" on the left and "3 Fe(s)" on the right. The given equation that relates those two species is the second one. So multiply the second given equation by 3/2, then write it backwards:
3 Fe(s) + 9/2 CO2(g) → 3/2 Fe2O3(s) + 9/2 CO(g), ΔH° = +34.5 kJ
Add the last two equations here:
3 FeO(s) + 7/2 CO(g) + 3/2 Fe2O3(s) + 3 Fe(s) + 9/2 CO2(g) →
3 Fe(s) + 7/2 CO2(g) + Fe3O4(s) + 3/2 Fe2O3(s) + 9/2 CO(g), ΔH° = -52.5 kJ +34.5 kJ
Cancel like amounts on opposite sides of the arrow, and do the arithmetic for ΔH°:
3 FeO(s) + CO2(g) → Fe3O4(s) + CO(g), ΔH° = -18 kJ

[It would be a good idea to check all my arithmetic concerning ΔH°. I would have guessed the final answer to be positive, so I'm not completely convinced my answer here is correct.]

Reverse the first given thermochemical equation and multiply (1/2) to the both sides :
(3/2)Fe₂O₃(s) + (1/2)CO(g) → Fe₃O₄(s) + (1/2)CO₂(g) ...... ΔH° = -(1/2)(+39.0 kJ)

Reverse the second given thermochemical equation and multiply (3/2) to the both sides :
3Fe(s) + (9/2)CO₂(g) → (3/2)Fe₂O₃(s) + (9/2)CO(g) ...... ΔH° = -(3/2)(-23.0 kJ)

Reverse the third given thermochemical equation and multiply 3 to the both sides :
3FeO(s) + 3CO(g) → 3Fe(s) + 3CO₂(g) ...... ΔH° = -3(+11 kJ)


Add the above three thermochemical equations, and cancel (3/2)Fe₂O₃(s), 3CO(g), 3CO₂(g) and 3Fe(s) on the both sides :
3FeO(s) + CO₂(g) → Fe₃O₄(s) + CO(g) ...... ΔH° = -18 kJ

[Calculations :]
[ΔH° = -(1/2)(+39.0 kJ) - (3/2)(-23.0 kJ) - 3(+11 kJ) = -18 kJ

42

What the fvck is this?