Two plants that are AaBbCc are crossed. The three genes lie on three different chromosomes, and…

Step 1. (a) $\left(\tfrac{1}{4}\right)^3 = \tfrac{1}{64} = 1.5625\%$.

Step 2. (b) $\left(\tfrac{3}{4}\right)^3 = \tfrac{27}{64} = 42.19\%$.

Step 3. (c) One gene is dominant and two are recessive. The chance is $\tfrac{3}{4} \times \tfrac{1}{4} \times \tfrac{1}{4} = \tfrac{3}{64}$, and there are $3$ choices of which gene is dominant: $3 \times \tfrac{3}{64} = \tfrac{9}{64} = 14.06\%$.