Part (b) The genotype of offspring number 1 is. Part (c) The probability of having a hemophiliac son is.

Offspring 1 is a male. His parents are the first generation male (normal) and female (carrier).

Hemophilia is X-linked recessive. Let $X^H$ be the allele for normal clotting and $X^h$ be the allele for hemophilia.
The normal male parent has genotype $X^H Y$.
The carrier female parent has genotype $X^H X^h$.

Offspring 1 is a normal male. Males inherit $Y$ from their father and an X chromosome from their mother. Since the father is $X^H Y$, he passes $Y$. The mother is $X^H X^h$. For the son to be normal, he must inherit $X^H$ from his mother.
Therefore, the genotype of offspring 1 is $X^H Y$.

The genotype of offspring number 1 is $\boxed{X^H Y}$.

A carrier woman has genotype $X^H X^h$.
A hemophiliac man has genotype $X^h Y$.

Gametes from woman: $X^H$, $X^h$
Gametes from man: $X^h$, $Y$

The possible genotypes for sons are $X^H Y$ (normal male) and $X^h Y$ (hemophiliac male).

Out of the four possible offspring genotypes, one is $X^h Y$, which represents a hemophiliac son.
The probability is $\frac{1}{4}$.

The probability that the couple will have a son who is hemophiliac is $\boxed{\frac{1}{4}}$.