Determination of spinel structure by CFSE
I'm not really getting your numbers here.
In your situation the iron is Fe(III), so a $d^5$ system. For $d^5$ it does not matter if it's in an octahedron or a tetrahedron since the energies will just cancel out. So in both situations it will be determined by the other ion. For Zn(II) it also doesn't really matter. So I guess here the regular spinel dominates as Zn(II) just prefers more tetrahedral coordinations in general, especially in solid states while Fe(III) is happy with an octahedron.
For the Cu(II) however you've got a $d^9$ system. So the octahedron will give you something like $(6*2/5)-(3*3/5) = 3/5$ (ignoring things like spin pairing energ here).
But in the tetrahedral case you also have to consider that the ligand field splitting for a tetrahedron is only $4/9$. So it will be something like $4/9*[(4*3/5)-(5*2/5)] = 8/45$ and therefore lower.
So the Cu(II) will prefer the octahedral position here and hence form an inverse spinel. Feel free to correct me here I was just drawing the diagrams as I read the question, perhaps I did a mistake somewhere.