Determine the father's genotype.

The man suffers from haemophilia, which is a sex-linked recessive condition. Since 'h' represents the allele for haemophilia, his genotype is $X^h Y$.

The woman is described as normal, but their daughter, Mercy, is haemophiliac. For Mercy to be haemophiliac, her genotype must be $X^h X^h$. She inherits one $X^h$ from her father ($X^h Y$) and one $X^h$ from her mother. Since the mother is phenotypically normal but carries the haemophilia allele, her genotype must be $X^H X^h$ (a carrier).

The genotype of the parents are:

• Father: $\boxed{X^h Y}$
• Mother: $\boxed{X^H X^h}$

Mother ($X^H X^h$): Gametes are $X^H$ and $X^h$.
Father ($X^h Y$): Gametes are $X^h$ and $Y$.

The possible genotypes of the offspring are:

• $X^H X^h$ (Carrier female, normal phenotype)
• $X^H Y$ (Normal male)
• $X^h X^h$ (Haemophiliac female)
• $X^h Y$ (Haemophiliac male)

From the Punnett square, the male offspring genotypes are $X^H Y$ (normal) and $X^h Y$ (haemophiliac).
There are 2 possible male genotypes, and 1 of them is haemophiliac.
The probability of a son being haemophiliac is $\frac{1}{2}$.
Probability of son being haemophiliac: $\boxed{\frac{1}{2} \text{ or } 50\%}$

From the Punnett square, the female offspring genotypes are $X^H X^h$ (carrier, normal phenotype) and $X^h X^h$ (haemophiliac).
There are 2 possible female genotypes, and 1 of them is haemophiliac.
The probability of a daughter being haemophiliac is $\frac{1}{2}$.
Probability of daughter being haemophiliac: $\boxed{\frac{1}{2} \text{ or } 50\%}$