Oxidation state of ions in a unit cell
This is a really weird question.

Let's start with the general case for a FCC lattice, you'd count up the oxygen atoms per unit cell. In the $\ce{NaCl}$ example, as noted in the comment, there are four $\ce{Cl-}$ per unit cell (6 on the faces, 8 on the corners). Then you'd count up the number of Al atoms per unit cell and divide up the charge as you note above.

In the case of "fractional" oxidation state, it implies that there's a mixture of oxidation states.

Let's say, hypothetically, you had 3 cations and need a +8 charge to balance out the anions. That tells me there are two +3 cations and a +3 cation somewhere. The average is +8/3 but I need to assign formal charges as integers.

But...

Normal $\ce{Al2O3}$ is not an FCC lattice. It's triclinic.



Here, you would do the same thing - count up the number of Al and O atoms per unit cell, considering the fractional volume of the atoms inside the unit cell. It's a bit trickier, since the volumes aren't as obvious as the FCC case.

Instead

This question is about $\gamma$-alumina



Notice that the oxygen atoms actually form an FCC lattice. I have an oxygen at each corner of the unit cell, and one in the middle of each face. So that's 4 oxygen atoms per unit cell = +8 charge from the anions.

How many Al atoms are there?



There are 5 Al atoms completely in the unit cell, plus one per edge (12, each counting 1/4 = 3) so a total of 8 Al atoms!

So that's an average oxidation state of +1 per Al...