Answer
$$2F_2(g) + 2H_2O (l) \longrightarrow 4 HF(g) + O_2(g) $$
$$\Delta H^o_{rxn} = -521 \ kJ/mol$$
Work Step by Step
1. Multiply the first reaction by 2:
$$2H_2 (g) + 2 F_2 (g) \longrightarrow 4 HF (g)$$
$$\Delta H ^o _{rxn} = -546.6 \ kJ/mol * 2 = -1093 \ kJ/mol$$
2. Reverse the second equation:
$$2H_2O(l) \longrightarrow 2H_2(g)+O_2(g)$$
$$\Delta H^o_{rxn} = 571.6 \ kJ/mol$$
3. Sum these equations
$$2H_2(g) + 2F_2(g) + 2H_2O (l) \longrightarrow 4 HF(g) + 2 H_2(g) + O_2(g) $$
$$2F_2(g) + 2H_2O (l) \longrightarrow 4 HF(g) + O_2(g) $$
$$\Delta H^o_{rxn} = -1093 \ kJ/mol + 571.6 \ kJ/mol = -521 \ kJ/mol$$
