You can easily solve for the enthalpy change of reaction for your particlar reaction by manipulating the two given reactions.
color(blue)((1)): Ca_((s)) + CO_(2(g)) + "1/2"O_(2(g)) -> CaCO_(3(s)), DeltaH_1 = "-812.8 kJ"
color(blue)((2)): 2Ca_((s)) + O_(2(g)) -> 2CaO_((s)), DeltaH_2 = "-1269.8 kJ"
Now, think of how you can manipulate these two reactions to get to
CaO_((s)) + CO_(2(g)) -> CaCO_(3(s))
Notice that your target reaction has 1 mole of CaO reacting, while reaction color(blue)((2)) has 2 moles of CaO being produced.
If you flip equation color(blue)((2)) and divide it by 2, you'll get
color(blue)((2))/2 => Ca_((s)) + "1/2"O_(2(g)) -> CaO_((s))
(DeltaH_2)/2 = ("-1269.8 kJ")/2 = "-634.9 kJ"
If you flip it, the reverse reaction will be
color(blue)((2) "reversed") => CaO_((s)) -> Ca_((s)) + "1/2"O_(2(g))
DeltaH_("2 reversed") = "+634.9 kJ"
Now just add reaction color(blue)((1)) to reaction color(blue)((2) "reversed") to get
cancel(Ca_((s))) + CO_(2(g)) + cancel("1/2"O_(2(g))) -> CaCO_(3(s))
CaO_((s)) -> cancel(Ca_((s))) + cancel("1/2"O_(2(g)))
CaO_((s)) + CO_(2(g)) -> CaCO_(3(s))
Now do the same for the DeltaHs
DeltaH_"rxn" = DeltaH_1 + DeltaH_("2 reversed")
DeltaH_("rxn") = "-812.8 kJ" + "634.9 kJ" = color(green)("-177.9 kJ")
