Two people are carriers of the autosomal recessive allele for phenylketonuria…

Step 1. (a) $\left(\tfrac{3}{4}\right)^4 = \tfrac{81}{256} = 31.64\%$.

Step 2. (b) The affected child can be any one of the $4$. For one choice the chance is $\tfrac{1}{4} \times \left(\tfrac{3}{4}\right)^3 = \tfrac{27}{256}$. For four choices: $\tfrac{108}{256} = 42.19\%$.

Step 3. (c) At least one is the rest of the cases after none: $1 - \tfrac{81}{256} = \tfrac{175}{256} = 68.36\%$.