This is a one question at a time deal here. I'll do one of these and that should point you to the method of doing the others.
2 Fe³⁺(aq) + 3 Sn(s) → 2 Fe(s) + 3 Sn²⁺(aq)
First look up the half-cell potentials (I'm using those linked below):
There is no reduction potential in the list for the iron half cell reduction. However, there are two you can use to get it:
Fe³⁺ + e⁻ ⇌ Fe²⁺ . . . . . E⁰ = + 0.77 V . . . . (n = 1, nE⁰ = + 0.77 V)
Fe²⁺ + 2e⁻ ⇌ Fe(s) . . . E⁰ = – 0.44 V . . . . (n = 2, nE⁰ = – 0.88 V)
––––––––––––––––––––––––––––––––––––– . . . ––––––––––––––––––––
Fe³⁺ + 3e⁻ ⇌ Fe(s) . . . E⁰ = – 0.037 V . ⇐ . (n = 3, nE⁰ = – 0.11 V)
What I'm doing here is adding the nE⁰ 's, NOT the E⁰ 's. Remember that –nFE⁰ = ∆G⁰ so I'm adding up quantities that are directly proportional to free energies (I can always add up thermodynamic state function quantities – just like adding up ∆H's using Hess's Law.)
That was the tricky part, the rest is easy:
Sn²⁺ + 2e⁻ ⇌ Sn(s) . . . E⁰ = −0.13 V (The sign switches in the final reaction.)
2 Fe³⁺(aq) + 3 Sn(s) → 2 Fe(s) + 3 Sn²⁺(aq) . . . . E⁰(cell) = 0.13 – 0.037 V = 0.093 V ;
Note: n = 6 (total # of moles of electrons transferred in the reaction as written.)
∆G⁰(rxn) = –nFE⁰(cell) = –6(98.485 kJ/V)(0.093 V) = –54.9 kJ
Good luck on the other two examples.