It turns out that you can do very slightly better than the dice in your example. The probability has to be a multiple of $1/36$, your example has $20/36$, and the best possible value is $21/36=7/12$.
From my answer to the question that anon linked to, we know that the probability of one $6$-sided die beating another while having a lower median can be at most $3/4-1/12=2/3=24/36$. Wikipedia gives an example of a cycle of four dice in which each die beats the next one with this probability.
You're looking for a cycle of three dice in which each die beats the next one with the same probability. Without loss of generality we can assume that the dice carry $18$ different numbers. Then there are $18!/6!^3=17153136$ different sets of dice, so it's well within the reach of our electronic friends to try them all out. Here's code that does that (rather inefficiently). The result is that the maximal probability for such a cycle is $21/36=7/12$ and there are eight different sets of dice that attain that probability. Here they are, with consecutive stretches of numbers collapsed into one as in your example:
0 3 3 3 6 6
2 2 2 5 5 5
1 4 4 4 4 4
0 0 3 3 3 6
2 2 2 2 2 5
1 1 1 4 4 4
0 3 5 5 5 5
2 4 4 4 4 7
1 1 1 6 6 6
0 3 5 7 7 7
2 2 4 4 6 9
1 1 1 8 8 8
0 3 3 5 5 5
2 2 2 4 4 7
1 1 1 6 6 6
0 3 5 5 7 7
2 2 2 4 6 9
1 1 1 8 8 8
0 3 3 3 3 5
2 2 2 2 4 7
1 1 1 6 6 6
0 0 3 6 6 6
2 2 2 5 5 5
1 1 1 4 7 7
There are two sets with $7$ different numbers, four sets with $8$ and two with $10$, and your near-optimal example has $9$, whereas the mean number of different numbers after collapsing, averaged over all sets with $18$ different numbers, is $18-17\cdot(5/17)=13$, so consecutive stretches of numbers and thus high collapsibility are a striking feature of these sets.
P.S.: I just realized that there's a symmetry in that you can invert the orders of both the dice and the numbers and get another set with the same probability. The first two sets are related by this symmetry, as are the third and seventh, the fourth and sixth and the fifth and eighth. Thus there are only four really different sets of dice, with $7$, $8$, $10$ and $8$ different numbers, respectively.
If you roll two dice the number of ways to get a total of seven is 6 as you could roll any of the following (1,6) (2,5) (3,4) (4,3) (5,2) (6,1). Since there are 36 possible results then from one roll the probability of total not being 7 is
$$\frac{36 - 6}{36} = \frac{30}{36} = \frac{5}{6}$$
For no six from two rolls the probability of no 7 is
$$\left(\frac{5}{6}\right)^2 = \frac{25}{36} \approx 0.6944$$
And for X rolls it's simply
$$\left(\frac{5}{6}\right)^X$$
Note: that because the outcome of a roll is random if you roll 6 times while counting the result then roll 3 times not counting the result (or even a million times) then roll another 14 times counting the result the probability is still
$$\left(\frac{5}{6}\right)^{20} \approx 0.026084$$
Best Answer
I am going to answer Part $2$ first, then part $1$. We can use the Binomial Formula here:
$$P(x) = {n \choose x} p^x q^{n-x}$$
Where assuming a fair die,
$n = 10$, the number of trials
$x=5$, the number of successes, i.e. $50 \%$ of $10$
$p = \dfrac 16$, the probability of success i.e. rolling a $5$ (or any other specific number)
$q = \dfrac 56$, the probability of failure, i.e. NOT rolling a $5$ (or any other specific number)
Plugging it all in, we get $\approx .01302$, or about $1.302 \% $.
It's probably more accurate to find the probability of getting a $5$ at least $50 \%$ of the time, because you would be equally or even more impressed with any success rate $\ge 50 \%$. To do this, you want to find the probability of $(x=5)$ OR $(x=6)$ OR $\cdots$ OR $(x=10)$, which is equal to $P(5) + P(6) + \cdots P(10)$. Plugging this in, we get $\approx .01546$, or about $1.546 \%$.
To find the probability of getting any $2$ numbers, Daugmented gives a nice idea in the comments. But we can still brute force our way through with the Binomial formula:
For getting $2$ numbers exactly $50$ of the time, we have
$n = 10$, the number of trials
$x=5$, the number of successes, i.e. $50 \%$ of $10$
$p = \dfrac 26 = \dfrac 13$, the probability of success, i.e. rolling a $5$ or a $6$ (or any other pair of numbers)
$q = \dfrac 46 = \dfrac 23$, the probability of failure, i.e. NOT rolling a $5$ or a $6$ (or any other pair of numbers)
The probability of getting a pair of numbers exactly $50 \%$ of the time is $\approx .1366$, or about $13.66 \%$.
Calculating as in part $2$ above, the probability of getting a pair at least $50 \%$ of the time is $\approx .2131$, or about $21.31 \%$.