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 $21.31 \%$.
When doing these kinds of computations in probability, it's often easier to consider the opposite event of what you're actually interested in. Here, the event you care about is "at least one die is a 6" (for example), so the opposite event is "no die is a 6." This means that die 1 is not a 6, and die 2 is not a 6, and dice 3, 4, 5 are all not a 6. The key in this computation is the word "and," because the independence of the dice means we can multiply those individual probabilities. The chance that no die is a 6 is therefore
$$\frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 = \frac{3125}{7776}$$
and the probability of the event you want is thus $ 1 - \frac{3125}{7776} = \frac{4651}{7776} \approx 0.5981$.
To answer your question a bit more comprehensively, you asked about when to use the addition rule and when to use the multiplication rule. At a basic level, multiplication corresponds to the word "and," and addition corresponds to the word "or." However, there are caveats; multiplication corresponds to "and" when events are independent, and addition corresponds to "or" when events are disjoint. Recognizing when these things happen is one of the main challenges of learning probability theory and comes only with a fair bit of experience in the subject, as far as I can tell.
In this case, the reason I knew to consider the complementary event was that the main event you wanted to consider had to be expressed in terms of several "or" events that were relatively complicated (i.e. "exactly 1 die is a 6," "exactly 2 dice are a 6," etc.) but the opposite event was not complicated in that way.
Best Answer
Let's see how many ways we can obtain $5$ distinct numbers from $6$ rolls. First choose which $5$ numbers will appear in $\binom{6}{5}$ ways. Now observe that the only way to obtain $5$ distinct numbers in $6$ rolls is to have $4$ of the values appear exactly once, and one to appear exactly twice. There are $5$ choices for which value will appear twice and the value will appear in $\binom{6}{2}$ locations. Then the remaining $4$ rolls can be ordered in $4!$ ways. This gives a total of $$\binom{6}{5}\cdot 5\cdot\binom{6}{2}\cdot 4!$$ desirable rolls. There are $6^{6}$ possible rolls so the probability of obtaining a roll with $5$ distinct numbers is $$\frac{\binom{6}{5}\cdot 5\cdot\binom{6}{2}\cdot 4!}{6^{6}}$$
Now let's see how many ways we can obtain $4$ distinct numbers from $6$ rolls. First choose which $4$ numbers will appear in $\binom{6}{4}$ ways. Now there are two cases to consider: one value appears $3$ times and each of the other values appears once, or two values appear twice and two values appear once. We'll handle these separately.
Combining these cases, the probability of obtaining $4$ distinct numbers is $$\frac{\binom{6}{4}\left[4\cdot\binom{6}{3}\cdot 3! + \binom{4}{2}\cdot\binom{6}{2}\cdot\binom{4}{2}\cdot 2\right]}{6^{6}}$$