Identify the dominant allele.

The allele for brown eyes is dominant, represented by $E$.

A dominant phenotype can be homozygous dominant or heterozygous.
The two possible genotypes are:

• $EE$ (homozygous dominant)
• $Ee$ (heterozygous)

The two possible genotypes for a brown-eyed rabbit are $\boxed{EE \text{ and } Ee}$.

The allele for blue eyes is recessive, represented by $e$.

A recessive phenotype must be homozygous recessive.

The only possible genotype for a blue-eyed rabbit is $\boxed{ee}$.

Homozygous brown-eyed rabbit: $EE$
Blue-eyed rabbit: $ee$

Parent 1 ($EE$): $E$
Parent 2 ($ee$): $e$

All offspring will have the genotype $Ee$.
All offspring will have brown eyes.

The genetic diagram shows that all offspring will be $\boxed{\text{brown-eyed}}$.

Heterozygous brown-eyed rabbit: $Ee$
Blue-eyed rabbit: $ee$

Parent 1 ($Ee$): $E, e$
Parent 2 ($ee$): $e$

Genotypes: $Ee$ (50%), $ee$ (50%)
Phenotypes: Brown eyes (50%), Blue eyes (50%)

The genetic diagram shows that the expected phenotypes are $\boxed{50\% \text{ brown-eyed and } 50\% \text{ blue-eyed}}$.

Blue-eyed parent: $ee$
Offspring: 4 brown-eyed, 5 blue-eyed (approximately a 1:1 ratio).

Case 1: If the brown-eyed parent was $EE$.
Cross: $EE \times ee$
Offspring: All $Ee$ (brown-eyed). This does not match the observed offspring (some blue-eyed).
Case 2: If the brown-eyed parent was $Ee$.
Cross: $Ee \times ee$
Offspring: $50\%$ $Ee$ (brown-eyed) and $50\%$ $ee$ (blue-eyed). This is a 1:1 ratio.

The observed ratio of 4 brown-eyed to 5 blue-eyed offspring is approximately 1:1, which matches the expected ratio from an $Ee \times ee$ cross.

The genotype of the brown-eyed parent was $\boxed{Ee}$.

f) A rabbit breeder crossed a brown-eyed rabbit with a blue-eyed rabbit. The rabbits had three young, all of which had brown eyes. The breeder said this proved that the brown-eyed rabbit must have the genotype EE, but her daughter said she could not be certain and she should do the cross again to make sure. Who was right – the breeder or her daughter?

Blue-eyed rabbit: $ee$
Offspring: 3 young, all brown-eyed.

Case 1: If the brown-eyed parent was $EE$.
Cross: $EE \times ee$
Offspring: All $Ee$ (brown-eyed). This is consistent with observing 3 brown-eyed offspring.
Case 2: If the brown-eyed parent was $Ee$.
Cross: $Ee \times ee$
Offspring: $50\%$ $Ee$ (brown-eyed) and $50\%$ $ee$ (blue-eyed).

If the brown-eyed parent was $Ee$, there is a $\frac{1}{2}$ probability for each offspring to be brown-eyed. The probability of having 3 brown-eyed offspring in a row from an $Ee \times ee$ cross is $(\frac{1}{2})^3 = \frac{1}{8}$. This is a low probability, but it is not impossible.

Since there is still a chance (1 in 8) that an $Ee$ parent could produce 3 brown-eyed offspring, observing only 3 brown-eyed offspring does not conclusively prove that the brown-eyed parent is $EE$. More offspring would be needed to increase certainty.

Therefore, the $\boxed{\text{daughter}}$ was right.