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 $21.31 \%$.
There are 57 dice. If one is chosen at random (with uniform probability) then the probability of choosing a six-sided die and rolling a five is:
$\frac{36}{57} \frac{1}{6} = \frac{6}{57}$.
The probability of rolling a 5 in general is:
$\frac{13}{57} \frac{0}{4} + \frac{36}{57} \frac{1}{6} + \frac{8}{57} \frac{1}{8} = \frac{6}{57} + \frac{1}{57} = \frac{7}{57}$
So the conditional probability is:
$\frac{6}{57}$ divided by $\frac{7}{57}$, which is $\frac{6}{7}$.
6 and 7 are coprime, and $6+7=13$
Best Answer
You could argue completely with Bayes' theorem, but you can also observe that uniformly picking a die and rolling it once amounts to uniformly picking one of the $ 1200$ faces. You picked a six (out of $250$ sixes). There are $150$ sixes on "even" dice and $100$ sixes on normal dice (plus none on "odd" dice). Hence the probability that your particular six is on a normal die is $\frac{100}{250}$.