Answer: Aa is normal, and aa is Albinism.
1. 3 children
2. all female (circles are F, squares are M)
3. 1 grandchild
4. male
5. The child of that couple ended up having Albinism, which would only be possible if the wife was a carrier for the trait and passed it on.
6. I think there's three but it looks cut off, so...hope that's correct XD
7. two
Now for filling in the blanks:
Pedigree 1 -
Black square - X^aY ; Wife - X^AX^a
Children (left to right) - X^AX^A OR X^AX^a for the first two, and X^aX^a for the last daughter (black circle)
Pedigree 2 -
White square - X^AY ; Wife - X^aX^a
2nd generation (left to right) - X^aX^a, X^AY, (X^AY -> wife is X^AX^a)
Grandchild - X^aY
Pedigree 3 -
White square - X^AY ; Wife - X^AX^a
2nd generation (left to right) - X^aX^a, (X^AY -> wife is X^AX^A or X^AX^a)
I can't see the rest of the pedigree, but hopefully this helps.
Some notes: when both parents are normal and a male child ends up having the recessive trait, this means that the wife must be a carrier.
When the father has the trait and the mother doesn't and a daughter has the trait, the mother is a carrier as well.