Given the standard enthalpy changes for the following two reactions: (1) 2Pb(s) + O2(g)2PbO(s)...... ΔH° = -434.6 kJ (2) 2Hg(l) + O2(g)2Hg?

Rewrite the two given thermochemical equations as:
PbO(s) → Pb(s) + (1/2)O₂(g) …… ΔH° = -(1/2)(-434.6 kJ) = +217.3 kJ
Hg(l) + (1/2)O₂(g) → HgO(s) ...... ΔH° = (1/2)(-181.6 kJ) = -90.8 kJ

Add the above two thermochemical equation, and cancel (1/2)O₂(g) on the both sides:
PbO(s) + Hg(l) → Pb(s) + HgO(s) ...... ΔH° = (+217.3 - 90.8) kJ = +126.5 kJ