In the given question, there is a cross between III-6 and III-7 that produced two double eyes monsters and expected one more. As given in the image that the A or single eye is dominant over the a or double eye then both parents must be heterozygous as they produced double eye offspring (aa) and they do have only one eye.
then chances of single eye offspring would be -
cross: Aa and Aa
gametes: A, a and A, a
Punnett square:
A a
A AA Aa
a Aa aa
so there are 75% chances to have a single eye and 25% chances for a double eye.
1/4. Make a Punnett square and plug in the two heterozygous genotypes of the parents, put them together and you get the following possible genotypes: AA, Aa, Aa, and aa. One of those is homozygous dominant. 1/4 or 25 percent.