Hemophilia is an X-Linked recessive disorder. Cross the following parents and give the number of the

In the sex linked situation, the mother has two alleles and can be homozygous dominant (normal phenotype), heterozygous (carrier, normal phenotype) or homozygous recessive (affected phenotype). In contrast, the father has one allele (hemizygous) and the allele can be either the recessive allele (affected phenotype) or the dominant allele (unaffected phenotype.  So:

1.  The affected male's genotype is X(h)Y, the unaffected female's genotype is either X(H)X(H) or X(H)X(h).  We have to assume that the description of "normal" means that the female is homozygous dominant, X(H)X(H).

So, the cross is:  X(H)X(H)  x  X(h)Y
The offspring will be either X(H)X(h) carrier females or X(H)Y unaffected males.

You have the numbers correct in your answer (1/2 probability for each)

2.  Normal male X(H)Y  x X(H)X(h) carrier female
The offspring will be:
X(H)X(H) unaffected female
X(H)X(h) carrier female
X(H)Y unaffected male
X(h)Y affected male
All in equal proportions (1/4 for each)

3.  Normal male X(H)Y  x  X(h)X(h) affected female
The offspring will be:
X(H)X(h) carrier female
X(h)Y affected male
in equal proportions (1/2 for each)

4.  Affected male X(h)Y  x  X(H)X(h) carrier female
The offspring will be:
X(H)X(h) carrier female
X(h)X(h) affected female
X(H)Y unaffected male
X(h)Y affected male
All in equal proportions (1/4 for each)

X' = affected allele
X = normal allele

Just make a punnett square for each situation:

2. Normal male X Carrier female

XY x X'X

Normal daughter_____1/4__
Carrier daughter_____1/4__
Affected daughter____0/4__
Normal son___1/4____
Affected son___1/4___

3. Normal male X Affected female

XY x X'X'

Normal daughter_0/4____
Carrier daughter_2/4____
Affected daughter__0/4___
Normal son__0/4____
Affected son__2/4___

4. Affected male X Carrier female

X'Y x X'X

Normal daughter_0/4____
Carrier daughter_1/4_____
Affected daughter__1/4___
Normal son_1/4____
Affected son_1/4____