Once again, to get to the reaction of interest, all you have to do is manipulate the three given reactions.
color(blue)((1)): C_((s)) + 2H_(2(g)) -> CH_(4(g)), DeltaH_1 = "-74.6 kJ"
color(blue)((2)): C_((s)) + 2Cl_(2(g)) -> C Cl_(4(g)), DeltaH_2 = "-95.7 kJ"
color(blue)((3)): H_(2(g)) + Cl_(2(g)) -> 2HCl_((g)), DeltaH_3 = "-184.6 kJ"
Now look at your main reaction and try to determine what you have/need from the above three reactions
CH_(4(g)) + 4Cl_(2(g)) -> C Cl_(4(g)) + 4HCl_((g))
Notice that the main reaction has CH_4 on the reactants' side, while reaction color(blue)((1)) has it on the products' side -> flip reaction color(blue)((1)) to get CH_4 on the reactants' side
color(blue)((1) "reversed") => CH_(4(g)) -> C_((s)) + 2H_(2(g))
DeltaH_("1 reversed") = "+74.6 kJ"
Now notice that you need 4 moles of HCl for the main reaction, but only get 2 moles from reaction color(blue)((3)) -> multiply reaction color(blue)((3)) by 2 to get the correct number of HCl moles on the products' side
color(blue)((3)) * 2 => 2H_(2(g)) + 2Cl_(2(g)) -> 4HCl_((g))
DeltaH_("3 doubled") = 2 * DeltaH_3 = 2 * ("-184.6 kJ") = "-369.2 kJ"
Now add all the three resulting reactions to get your main one,
color(blue)("(1) reversed") + color(blue)((2)) + color(blue)(("3 doubled"))
CH_(4(g)) -> cancel(C_((s))) + cancel(2H_(2(g)))
cancel(C_((s))) + 2Cl_(2(g)) -> C Cl_(4(g))
cancel(2H_(2(g))) + 2Cl_(2(g)) -> 4HCl_((g))
CH_(4(g)) + 2Cl_(2(g)) + 2Cl_(2(g)) -> C Cl_(4(g)) + 4HCl_((g)), which is equivalent to
CH_(4(g)) + 4Cl_(2(g)) -> C Cl_(4(g)) + 4HCl_((g))
Do the same for the DeltaHs to get
DeltaH_("rxn") = DeltaH_("1 reversed") + DeltaH_2 + DeltaH_("3 doubled")
DeltaH_("rxn") = "+74.6 kJ" + ("-95.7 kJ") + ("-369.2 kJ")
DeltaH_("rxn") = color(green)("-390.3 kJ")