Principles of Inheritance and Variation

Genotype of the man: Aa Bb $X^hY$ (heterozygous for A and B, and hemizygous for h, i.e. one $X^h$ and a Y).
Assuming independent assortment, each gene pair separates independently into the sperms.
Probability of a sperm getting a $= \dfrac{1}{2}$ (A or a)
Probability of getting b $= \dfrac{1}{2}$ (B or b)
Probability of getting the X with h $= \dfrac{1}{2}$ (the other half of the sperms carry Y)
Probability of an abh sperm $= \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$
The man forms $2^3 = 8$ types of sperms in equal proportions (ABh, ABY, Abh, AbY, aBh, aBY, abh and abY), and abh is one of them.
So $\dfrac{1}{8}$ of his sperms will be abh.

A trihybrid plant is heterozygous for three genes, e.g. AaBbCc.
Number of types of gametes $= 2^n$, where n is the number of heterozygous gene pairs: $2^3 = 8$ (ABC, ABc, AbC, Abc, aBC, aBc, abC, abc).
On self-fertilisation, each of the 8 types of male gametes can fuse with each of the 8 types of female gametes.
Number of zygote combinations $= 8 \times 8 = 64$ (the squares of the Punnett square).
(These give 27 different genotypes and 8 phenotypes in the ratio 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1.)
So the plant forms 8 different gametes and 64 different zygotes.

Gametes of RRTt: RT and Rt (in equal proportion, 1/2 each).
Gametes of rrtt: rt only.
Offspring: RT $\times$ rt = RrTt (red, tall); Rt $\times$ rt = Rrtt (red, short).
So all offspring have red fruit (Rr), because the RRTt parent is homozygous for red.
For height, Tt $\times$ tt gives 1/2 Tt (tall) and 1/2 tt (short).
Ratio: 1 red tall : 1 red short.
So 50% of the offspring will be tall with red fruit.

Mitochondrial DNA is inherited through the cytoplasm.
During fertilisation, the egg contributes almost all of the cytoplasm (and mitochondria) to the zygote. The sperm contributes mainly its nucleus; its few mitochondria are lost or destroyed.
So mitochondrial traits show maternal (uniparental) inheritance: they are passed only through the female parent.
Here the parent with the mitochondrial mutation is the male, so its mitochondria are not transmitted to the $F_1$, and therefore not to the $F_2$ either.
So the mutation is found in none of the progenies.