How many outcomes there are depends on what you’re counting as different outcomes. If each possible choice of dice and each possible way the two chosen dice can come up is counted as a separate outcome, there are $2840$ possible outcomes; here’s the reasoning.
If you pick the $20$- and $12$-sided dice, there are $20\cdot12=240$ possible outcomes; if you pick the $20$-and $10$-sided dice, there are $20\cdot10=200$ possible outcomes; and so on. The total possible number of outcomes is therefore the sum of all possible products of two of the numbers $20,12,10,8,6$, and $4$. You can compute this by brute force, or you can take advantage of the fact that $$\begin{align*}&(a_1+a_2+a_3+a_4+a_5+a_6)^2=\\
&a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+a_6^2+2(\text{the sum that you want})\;,
\end{align*}$$ so that $$\begin{align*}&\text{the sum that you want}=\\
&(20+12+10+8+6+4)^2-(20^2+12^2+10^2+8^2+6^2+4^2)=\\
&60^2-(400+144+100+64+36+16)=\\
&3600-760=\\
&2840\;.
\end{align*}$$
These $2840$ outcomes are not all equally likely, however. Since there are $\binom62=15$ possible pairs of dice, and all pairs are equally likely, the probability of drawing any particular pair is $\frac1{15}$. If you draw the $20$- and $12$-sided dice, each of the $240$ possible rolls is equally likely, so the overall probability of drawing those two dice and getting a particular roll is $\frac1{15}\cdot\frac1{240}=\frac1{3600}$. If you draw the $6$- and $4$-sided dice, on the other hand, there are only $24$ possible rolls, so the probability that you draw those two dice and get a particular roll is $\frac1{15}\cdot\frac1{24}=\frac1{360}$, or ten times as large.
However, you might decide to count as different outcomes only the different possible totals on the two dice. The minimum possible total is $1+1=2$, the maximum is $20+12=32$, and every integer total between these extremes is possible, so there are $32-1=31$ possible totals, as you said. Of course these $31$ outcomes also have very different probabilities: to take the easiest example, there is only one way to get a total of $32$, but there are $15$ ways to get a total of $2$, one for every possible pair of dice.
This is relevant to the second question: because the different totals have different probabilities, you can’t simply say that there are $16$ even and $15$ odd totals, and therefore the probability of getting an even total is $\frac{16}{31}$. You’re going to have to analyze the possibilities in more detail.
Let’s look at a particular pair of dice, say the $10$- and the $6$-sided dice. The $10$-sided die has $5$ even and $5$ odd numbers, and the $6$-sided die has $3$ even and $3$ odd numbers. The $10\cdot 6=60$ possible rolls of these two dice can therefore be broken down as follows: $5\cdot 3=15$ rolls with an even number on both dice; $5\cdot 3=15$ rolls with an even number on the $10$-sided die and an odd number on the $6$-sided die; $5\cdot 3=15$ rolls with an odd number on the $10$-sided die and an even number on the $6$-sided die; and $5\cdot 3=15$ rolls with an odd number on both dice. When do you get an even total? In the first and last cases, which amount to $15+15=30$ out of the $60$, or half of them. Thus, if you happen to draw the $10$- and the $6$-sided dice, your probability of rolling an even total is $\frac12$.
We could have got this with much less work by realizing right away that in order to get an even total, we must get either two even or two odd numbers. Each die has half its faces even and half of them odd, so no matter which two dice we pick, the probability of getting even + even is $\frac12\cdot\frac12=\frac14$, and so is the probability of getting odd + odd, which means that the probability of getting an even total is $\frac14+\frac14=\frac12$. Since the probability of getting an even total is $\frac12$ no matter which pair of dice we draw, the overall probability of getting an even total must also be $\frac12$.
For the last question you must also look a bit more closely at the possible outcomes. First, you can’t roll an $8$ or higher on both dice unless both dice have at least $8$ faces, so you must begin by drawing two of the four dice with the most faces; there are $\binom42=6$ ways to do this, so $\frac9{15}=\frac35$ of the time you won’t even draw dice on which it’s possible to get $8$ or higher on both. What you should do is look at these six pairs individually, counting the number of rolls that give you two numbers that are at least $8$. For instance, if you’re rolling the $12$- and $10$-sided dice, you need to get one of the $5$ largest numbers on the $12$-sided die and one of the $3$ largest on the $10$-sided die; the probability of doing so is $\frac5{12}\cdot\frac3{10}=\frac18$. Of course the probability of drawing these two dice in the first place is only $\frac1{15}$, so the probability of drawing them and getting a successful roll is $\frac1{15}\cdot\frac18=\frac1{120}$. Similarly, the probability of success when you roll the $12$- and the $8$-sided dice is $\frac5{12}\cdot\frac18=\frac5{96}$, and the probability of getting those two dice in the first place is $\frac1{15}$, so the probability of drawing them and getting a successful roll is $\frac1{15}\cdot\frac5{96}=\frac1{288}$. If you perform similar calculations for each of the other four pairs of dice that can give you successful outcomes, you can then add the results to get the final probability of rolling two numbers greater than or equal to $8$.
You can’t simply count successful outcomes and divide by $2840$, since the $2840$ possible rolls aren’t equally likely, as noted at the beginning.
The probability that all sides are the same is $6$ times the probability that all sides are $1$. So we don’t have to worry about the distribution of all numbers on the die, we just have to keep track of the number of $1$s. That yields a Markov chain with $7$ different states, two of which ($0$ and $6$) are absorbing. The transition matrix is
$$
\mathsf P(i\to j)=6^{-6}\binom6ji^j(6-i)^{6-j}
$$
(where $0^0=1$), or in matrix form:
$$
P=6^{-6}\pmatrix{
46656&0&0&0&0&0&0\\
15625&18750&9375&2500&375&30&1\\
4096&12288&15360&10240&3840&768&64\\
729&4374&10935&14580&10935&4374&729\\
64&768&3840&10240&15360&12288&4096\\
1&30&375&2500&9375&18750&15625\\
0&0&0&0&0&0&46656\\
}\;.
$$
To my surprise, this matrix has a rather nice eigensystem:
$$
P=\pmatrix{6\\5&5&-5&25&-5&3725&1\\4&8&-4&-8&8&-10576&2\\3&9&0&-27&0&14337&3\\2&8&4&-8&-8&-10576&4\\1&5&5&25&5&3725&5\\&&&&&&6}\\
\times\pmatrix{6\\&38160\\&&120\\&&&2448\\&&&&120\\&&&&&648720\\&&&&&&6}^{-1}
\\
\times
\pmatrix{1\\&\frac56\\&&\frac59\\&&&\frac5{18}\\&&&&\frac5{54}\\&&&&&\frac5{324}\\&&&&&&1}
\\\times\pmatrix{1\\-2681&981&1125&1150&1125&981&-2681\\7&-8&-5&0&5&8&-7\\-14&33&-6&-26&-6&33&-14\\1&-4&5&0&-5&4&-1\\-1&6&-15&20&-15&6&-1\\&&&&&&1}\;,
$$
where the first diagonal matrix is just for normalization (so that I could write the matrices containing the left and right eigenvectors with integers) and the second diagonal matrix contains the eigenvalues.
Thus, since we start in state $1$, the probability to have reached state $6$ after $n$ rolls is
$$
-\frac{5\cdot2681}{38160}\left(\frac56\right)^n+\frac{5\cdot7}{120}\left(\frac59\right)^n-\frac{25\cdot14}{2448}\left(\frac5{18}\right)^n+\frac{5\cdot1}{120}\left(\frac5{54}\right)^n-\frac{3725\cdot1}{648720}\left(\frac5{324}\right)^n+\frac16
\\[3pt]
=
-\frac{2681}{7632}\left(\frac56\right)^n+\frac7{24}\left(\frac59\right)^n-\frac{175}{1224}\left(\frac5{18}\right)^n+\frac1{24}\left(\frac5{54}\right)^n-\frac{745}{129744}\left(\frac5{324}\right)^n+\frac16\;.
$$
We need to multiply this by $6$ to get the probability of having reached this state for any of the $6$ numbers on the die in order to get the probablity that the number $N$ of required rolls is less than or equal to $n$:
$$
\mathsf P(N\le n)=1-\frac{2681}{1272}\left(\frac56\right)^n+\frac74\left(\frac59\right)^n-\frac{175}{204}\left(\frac5{18}\right)^n+\frac14\left(\frac5{54}\right)^n-\frac{745}{21624}\left(\frac5{324}\right)^n\;.
$$
Here we can check that $\mathsf P(N\le0)=0$ and $\mathsf P(N\le1)=6^{-5}$, as they must be. The probability that we need exactly $n$ rolls is
\begin{eqnarray}
\mathsf P(N=n)
&=&
\mathsf P(N\le n)-\mathsf P(N\le n-1)
\\[3pt]
&=&
\frac{2681}{6360}\left(\frac56\right)^n-\frac75\left(\frac59\right)^n+\frac{455}{204}\left(\frac5{18}\right)^n-\frac{49}{20}\left(\frac5{54}\right)^n+\frac{47531}{21624}\left(\frac5{324}\right)^n
\end{eqnarray}
for $n\gt0$ and $\mathsf P(N=0)=0$. Here’s a plot, in agreement with your numerical results. The expected value of $N$ is
\begin{eqnarray}
\mathsf E[N]
&=&\sum_{n=0}^\infty\mathsf P(N\gt n)
\\[3pt]
&=&
\sum_{n=0}^\infty\left(\frac{2681}{1272}\left(\frac56\right)^n-\frac74\left(\frac59\right)^n+\frac{175}{204}\left(\frac5{18}\right)^n-\frac14\left(\frac5{54}\right)^n+\frac{745}{21624}\left(\frac5{324}\right)^n\right)
\\[6pt]
&=&
\frac{2681}{1272}\cdot\frac6{6-5}-\frac74\cdot\frac9{9-5}+\frac{175}{204}\cdot\frac{18}{18-5}-\frac14\cdot\frac{54}{54-5}+\frac{745}{21624}\cdot\frac{324}{324-5}
\\[6pt]
&=&
\frac{31394023}{3251248}
\\[6pt]
&\approx&9.656\;,
\end{eqnarray}
also in agreement with your numerical results.
Best Answer
Use a generating function for the dice. Normal dice have faces with 1 to 6 dots, so they are represented by the polynomial:
$\begin{align} D(z) &= z + z^2 + z^3 + z^4 + z^5 + z^6 \\ &= z \frac{1 - z^6}{1 - z} \end{align}$
Throwing two dice gives the distribution:
$\begin{align} D^2(z) &= z^2 + 2 z^3 + 3 z^4 + 4 z^5 + 5 z^6 + 6 z^7 + 5 z^8 + 4 z^9 + 3 z^{10} + 2 z^{11} + z^{12} \end{align}$
We want polynomials $D_1(z)$ and $D_2(z)$ that give:
$\begin{align} D_1(z) D_2(z) = D^2(z) \end{align}$
It is clear that $D_i(1)$ is the number of faces of the die, so we want:
$\begin{align} D_1(1) = D_2(1) = 6 \end{align}$
If we want no blank faces, it means $D_i(0) = 0$ (no constant term).
Let's factor:
$\begin{align} D(z) &= z (z + 1) (z^2 - z + 1) (z^2 + z + 1) \end{align}$
If we evaluate the factors at $z = 1$:
$\begin{align} D(1) &= 1 \cdot 2 \cdot 1 \cdot 3 \end{align}$
Thus, in order to have $D_i(1) = 6$, each has to get a factor $z + 1$ and a factor $z^2 + z + 1$. As we want $D_i(0) = 0$, each one has to get a factor $z$ too. This leaves two factors $z^2 - z + 1$ (one from each normal die) to distribute, giving the dice:
$\begin{align} D_1(z) &= z (z + 1) (z^2 - z + 1)^2 (z^2 + z + 1) \\ &= z + 2 z^2 + 2 z^3 + z^4 \\ D_2(z) &= z (z + 1) (z^2 + z + 1) \\ &= z + z^3 + z^4 + z^5 + z^6 + z^8 \end{align}$
These is dice with faces $\{1, 2, 2, 3, 3, 4\}$ and $\{1, 3, 4, 5, 6, 8\}$ The only other solution is normal dice (and exchanging them, obviously).
They are called Sicherman dice.