If A and B have kids, what's the probability of first child having PKU?

Answer = P(A is het) X P (B is het) X P (if both hets, both pass on recessive trait)

P(A is het) = P (A's father is het) X P (of passing recessive allele to A) = 1/2 X 1/2 = 1/4

P(B is het) = P (B's mother is het) X P (of passing recessive allele to B) = 2/3 X 1/2 = 2/6 = 1/3

Answer = P(A is het) X P (B is het) X P (if both hets both pass on recessive trait) = 1/4 X 1/3 X 1/4 = 1/48

If first kid is unaffected, what's the probability of second child having PKU?

Answer = the same. Nothing has changed in terms of probability.

If first kid is affected, what's the probability of second child will not have PKU?

The fact that the first kid had PKU tells you both parents were heterozygous, thus the chance of a normal child is the chance of the PP or Pp genotype = 3/4.