Answer:
A Punnett square for the following monohybrid cross can be shown as follows:
s s
S Ss Ss
S Ss Ss
a) The genotype of the potential offspring will be Ss, they will be heterozygous dominant.
b) There is a zero% probability that the offspring will have no freckles. As one of the parents was homozygous dominant for the freckles, hence all of the offsprings will have freckles.