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 $21.31 \%$.
