calculate standard enthalpy change for reaction?

You need "2 H2S(g)" on the left.  The only given equation that mentions H2S(g) is the first one, so write the first given equation backwards:
2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(g), ΔH = −146.9 kJ
You also need "3 O2(g)" on the left. The only given equation that mentions O2(g) is the second one, so multiply the second given equation by 3:
3 S(s) + 3 O2(g) → 3 SO2(g), ΔH = −889.2 kJ
Add the two equations here:
2 H2S(g) + SO2(g) + 3 S(s) + 3 O2(g) →3 S(s) + 2 H2O(g) + 3 SO2(g),
ΔH = −146.9 kJ −889.2 kJ
Cancel like amounts on opposite sides of the arrow, and do the arithmetic for ΔH:
2 H2S(g) + 3 O2(g) → 2 H2O(g) + 2 SO2(g), ΔH = −1036.1 kJ

Rewrite the two thermochemical equations as follows:
2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(g) … ΔH°₁ = -146.9 kJ/mol
3S(s) + 3O₂(g) → 3SO₂(g) … ΔH°₂ = 3(-296.4) = -889.2 kJ/mol

Add the above two thermochemical equations, and cancel 3S(s) and SO₂(g) on each side:
2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(g) … ΔH°
ΔH° = ΔH°₁ + ΔH°₂ = (-146.9) + (-889.2) = -1036.1 kJ/mol

Anon don't get no love....