A woman's brother has haemophilia. Their father and mother are healthy…

Step 1. (a) The woman gets one X from her mother, which carries h with a chance of $\tfrac{1}{2}$. So she is a carrier with a chance of $50\%$.

Step 2. (b) The two cases. Carrier with two healthy sons: $\tfrac{1}{2} \times \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{8}$. Not a carrier with two healthy sons: $\tfrac{1}{2} \times 1 = \tfrac{1}{2}$. The chance of being a carrier is $\dfrac{1/8}{1/8 + 1/2} = \dfrac{1}{5} = 20\%$.

Step 3. (c) For the next child to be a haemophilic son, she must be a carrier ($\tfrac{1}{5}$), and then the chance of a haemophilic son is $\tfrac{1}{4}$. So $\tfrac{1}{5} \times \tfrac{1}{4} = \tfrac{1}{20} = 5\%$.