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 #DeltaH#s 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")#
