Bred back to a cuckoo that doesn't carry mottled you would get cuckoo(barred) chicks with a 25% chance of carrying mottled.
Math error in this bit.
If one parent is carrying mottling (not showing it) and the other has no mottling gene at all, you should get 50% of chicks carrying mottling (not showing it), and the other 50% of chicks have no mottling gene.
This is because the parent who carries mottling has a 50/50 chance of passing that gene to each chick, and of course the parent with no mottling gene cannot pass it on at all.
Bred to a black mottled you should get 100% mottled chicks
Math error in this bit too:
If one parent carries mottling without showing it, and the other parent does show mottling, then you should get 50% mottled chicks and 50% chicks that carry mottling without showing it.
Again, the parent who is carrying mottling has a 50/50 chance of passing that gene to each chick. The parent who shows mottling will give that gene to every chick, so every chick has one mottling gene, but only the chicks who inherit a mottling gene from the carrier-parent will actually show the trait.
...with 50% also being barred. If you used a barred female all of the barred and mottled chicks would be male. If you use a barred male then barring would not indicate gender because a single barred male has a 50/50 chance of passing on barring to both male and female offspring.
I agree about the inheritance of barring here.
Combining that with the mottling numbers, it will be:
25% show mottling and barring
25% show mottling but not barring
25% carry mottling (not showing) and have barring
25% carry mottling (not showing) and have no barring
I agree about the ones showing barring being all male (mother had barring) or some of both sexes (father had barring).