A woman is healthy. Her parents are healthy, and her brother has an autosomal recessive condition…

(a) The woman is $AA$, $Aa$ or $Aa$ with weights $1 : 2$ among the three healthy outcomes of $Aa \times Aa$ (the fourth, $aa$, is ruled out). So $P(\text{carrier}) = \tfrac{2}{3} = 0.6667$.

(b) $2q / (1 + q) = (2/60) / (61/60) = \tfrac{2}{61} = 0.0328$.

(c) Both must be carriers, probability $0.6667 \times 0.0328 = 0.0219$, and then $\tfrac{1}{4}$ of children are affected. That is $54.64$ in $10000$.

(d) If both are carriers, a child is healthy with probability $\tfrac{3}{4}$, so two healthy children have probability $0.5625$. In every other state the children are surely healthy (probability $1$). By Bayes' rule, $P(\text{both carriers}) = 0.0219 \times 0.5625 / (0.0219 \times 0.5625 + 1 - 0.0219) = 0.0124$.

(e) The third child is affected only if both are carriers and then with probability $\tfrac{1}{4}$: $0.0124 / 4$, which is $31.03$ in $10000$. It is lower than the first child's $54.64$ because two healthy children are some evidence that the parents are not both carriers.