The pedigree shows a family with haemophilia…

Step 1. (a) The daughters of $X^H X^h$ and $X^H Y$ are $X^H X^H$ and $X^H X^h$, in equal numbers. II-1 is normal, which both are, so the chance that she is a carrier is $50\%$.

Step 2. (b) For a haemophilic son she must be a carrier, which has a chance of $\tfrac{1}{2}$. A carrier mother and a normal father have a haemophilic son with a chance of $\tfrac{1}{4}$ for each child. Together: $\tfrac{1}{2} \times \tfrac{1}{4} = \tfrac{1}{8} = 12.5\%$.

Step 3. (c) A haemophilic son can come only from a carrier mother, so II-1 is now certainly a carrier. The chance for each child is $\tfrac{1}{4} = 25\%$.