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 #DeltaH#s
#DeltaH_"rxn" = DeltaH_1 + DeltaH_("2 reversed")#
#DeltaH_("rxn") = "-812.8 kJ" + "634.9 kJ" = color(green)("-177.9 kJ")#
