What do you know? Then figure out the unknowns. "Autosomal" is the normal type of gene that you learned about first--it isn't carried on the sex chromosomes, it is on one of the other chromosomes.
From sentence three of the first paragraph, we know: D-dwarf d-normal B-normal b-colorblind (dominant is capitalized)
Here's how I started, more is explained below:
Grandfather--Father ("the man")'s father--dd
Father ("the man")-- D? By
Mother ("the woman")-- dd bb (you know exactly what she is contributing to the children--both traits she expresses are recessive)
#3 Daughter-- D? B? -- however, since you know what the mother contributed, you can eliminate those ?'s (more on that later).
1) Daughters get an X chromosome from each parent, but the father only has one X to express the trait and send to his offspring, while the mother has two X chromosomes. You know that the father's X chromosome does not carry the recessive gene, since he only has one X chromosome and he is not color blind . Daughters of a normal-sighted man have no chance of being colorblind--they always get the dominant gene from their father, so the gene from the mother is irrelevant to the phenotype.
2) Regarding color-blindness, Sons get an X from mom and a y from dad. This means that all of their sons will be genotype by and color-blind (sons of a colorblind mother are always colorblind--she only has colorblind genes to offer, and the father is not contributing an X chromosome). Regarding height, they will get a normal d from their mother. From their father, you know that there is at least a 50% chance of getting the D dwarf gene--he has to have at least one. Because the grandfather is normal height, you can deduce that father got the dwarf gene from grandmother, who is a dwarf, and (more importantly for this question) that his father gave him the recessive normal gene. Since we now know that father's genotype is Dd, sons have a 50% chance of being dwarves and 50% normal height. Then you just multiply (or think about) those results to get the combined answer: 100% chance of colorblindness times 50% chance of dwarfism equals 50% chance of a son being colorblind and of normal height.
3) The chance of her being heterozygous for both genes is 100%. We know what the mother is giving her daughter, as she is homozygous--she is giving db. We know that the daughter is D?B? because of her phenotype/description--if she didn't have those dominant genes, she couldn't be a normal-sighted dwarf. Since we know what the mother is contributing, and know what we know about the daughter, the daughter must be DdBb--heterozygous for both genes. Her father gave her both the normal vision and dwarfism.
Punnett squares don't help me, but if they help you, set up separate ones for each gene since they are on different chromosomes--don't attempt a big 16 square punnet square. Draw two squares (a D square and a B square) for each person who's genotype you are trying to determine. Remember that for the B square, males don't have a second B or b, they have a y instead).
While it is an oversimplification that isn't always true, dominant genes usually create a protein/enzyme that does something (good bad or just different), while the recessive gene lacks the code for making that protein. The father produces enzymes that stunt his growth and enzymes to help make the special parts of the eye that allows his eyes to see color. If someone lacks the code to make those enzymes, they have no enzyme to stunt their growth (normal height) and had no enzyme to make their eyes create the parts that see color.