Amale and a female mouse are each from pure-breeding albino strains. they have a litter of 10 pups, all of which have normal pigmentation. the f1 pups are crossed to one another to produce 56f2 mice, of which 31 are normally pigmented and 25 are albino.

You are forgetting the question but I can help you understand the exercise.

For this exercise it's important to know that there are more than one gene that causes albinism. In this exercise, one of the parents is albino because of one gene( gene A) and the other parent is albino because of another gene( gene B)
Albinism only happens if it's inherited both recessive allelles.
The male and female mouse follow those rules, so the cross between them is:
AAbb x aaBB

All their pups, the generation F1, is:AaBb
Since the disease is recessive (that means: aa), they don't have albinism.

After they cross each other: AaBb x AaBb (see the image attached)