The tricky part of this question is that colorblindness (red-green) is
recessive on the X chromosome, making it sex linked. See this page
for a discussion of the genetcs:
I'm calling our colorblind gene C and c, with the Y chromosome not
having either of them so noted just as Y.
Here are our parents' genetics:
Woman: BOCc (she is heterozygous because her father, who was cY, gave
her his colorblind X gene, but she's not colorblind so her other X
must have the dominant C)
In this case, since we have three different options for each gene
(A,B,O for blood; C, c, Y for colorblindedness), there are sixteen
different possible outcomes - which distinguishes us from our previous
questions. Usually you'll have two possible ways to get a
heterozygous child - but in this case each possible combination has a
1/16th chance of showing up.
I.e., you have 1/16th chance of having a colorblind boy with type A blood.
If it helps, you have 1/2 chance of getting a boy. There are four
blood types possible (AB, BB, BO, and AO), so you have a 1/4th chance
of type A blood. Half of your boys will be colorblind. Multiple the
probabilities: 1/2 times 1/4 times 1/2, and you get a 1/16th chance of
boy - AO - colorblind.