Wait! I think we cheated, we're probably not allowed to just hunt down the reaction eqn.
Ok, I think I know how to do it, I'll get you started:
Calculate quantity of O in solid product:
0.97g HF= 0.0485mol HF formed
0.0485/2= 0.02425mol H20 reacts (Since other product has no H)
From above, 0.02425mol of O in solid product (only product with O)
From the amount of UF6 reacted, you can work out how much U is in the solid product since other product has no U.
n(UF6)=4.267/352=0.01212mol
Given that you have 3.73g of solid product you can now solve for the quantity of F in the solid (you could also use the amount of HF formed and the amount of UF6 reacted).
b), c) and d) can be solved from here.
Hope that Helps