what are the concentrations (M) of Fe3+ and I- obtained by dissolving .200 mol FeI2 in water and diluting to 725ml?

It should be "Fe²⁺" istead of "Fe³⁺" because only FeI₂ is provided. Besides, Fe³⁺ ion does not exist in the presence of I⁻ because of redox reaction occurs.

Molarity of FeI₂ = (0.200 mol) / (725/1000 L) = 0.276 M

1 mole of FeI₂ dissociates in water to give 1 mole of Fe²⁺ ions and 2 moles of I⁻ ions.
Molarity of Fe²⁺ ion = (0.276 M) × 1 = 0.276 M
Molarity of I⁻ ion = (0.276 M) × 2 = 0.552 M