Well, I am not sure about retinitis pigmentosa, but the determination of ABO blood group is controlled by an autosomal codominant allele, meaning the blood group gene inherited from each parent will be expressed equally in the child. The genes for A and B code for substances which are added to the red cell membrane when they are produced in the bone marrow. Note that the O gene will code for nothing, while A and B will code for sugars on the red cell membrane.
A. Maria has group B blood, so her father had to be A/O, since we know his phenotype was A and he had a B daughter. Maria had to inherit the B gene from mom, and the "silent" O from dad. So Maria is genotype B/O (no giggles).
B. Matthew must be A/O, since his phenotype is A, and his mother's phenotype is B, his mom has to have the genotype B/O.
C. Since Maria carries both the B and O genes, and Matthew carries the A and O genes, the combinations of possibilities for any child can be represented in a simple table. The genes going across are Matthew's, Maria's are displayed vertically, and the potential offspring's genotypes in the matrix:
A O
B AB BO
O AO OO
Therefore, there are four different possibilities of genotype for any offspring: AB, BO, AO, and OO. Corresponding phenotypes are AB, B, A, and O. In this case, there is a 25% chance of any child having any of the four possible blood groups. Ergo, a 25% chance of being phenotype(and genotype) group O.
Note: The term blood "type" is more correctly confined to the Rh (D) antigen (positive or negative) which did not figure in this scenario. For ABO, always use the term blood "group" and you will sound like a clinical laboratory scientist.
D. There is only a 6.25% chance that any two children will have group AB blood, and a 25% chance they will both be boys, so only a 1.56% chance they will both be group AB boys.
E. The odds are 50% the girl will be heterozygous for A, 50% heterozygous for B, and 50% she will be heterozygous for O.
