You should instruct anydice to roll a number custom dice, where the definition of the custom die includes a number of zeroes equal to the number of die-faces that do nothing, and a number of ones for number of faces which count as 'successful'. E.g. for 4d10 where dice coming up 6 or more count as successful and contribute to your final score (also known as rolling 4 dice against difficulty 6), use this:
output 4d{0,0,0,0,0,1,1,1,1,1}
This can be abbreviated by to a format Xd{0:a,1:b}, where 'X' is the number of dice, 'a' is the number of faces that give no successes, and 'b' is the number of faces that do give a success, such that a=(difficulty-1) and b=(11-difficulty). So for 4 dice at difficulty 6, it's written like this:
output 4d{0:5,1:5}
If instead you want to roll seven dice and wonder how many will come up 9-or-better, use this:
output 7d{0:8,1:2}
Now, in some systems, such as Storyteller / World of Darkness / Exalted, dice that come up with the '1' face subtract from your final score. If so, replace the first zero with a '-1'. Conversely, if dice that come up with the '10' face count as two successes, replace the final '1' with a '2', like this (when rolling 12 dice against difficulty 7):
output 12d{-1,0:5,1:3,2}
As a side note, while it's possible to roll normal d10s and give conditional counting instructions, use custom functions etc., it seems to often cause such a workload for the server that it times out without giving a result, particularly with high dice pools. The custom die format seems to avoid the timeout problem. So from now on I avoid the less efficient method and stick to the one I described above.
I've just been told that using the
output 7d(d10>=9)
format is safe (where 9 is the difficulty), and when I just tested it it worked OK, but at other times I've had timeouts with it. So it's for you to decide whether to use it for cases where you don't care about 1's and 10's.
Reading the results
Anydice provides table and graph result formats, and can offer you an 'exactly' (probability of a given specific outcome), 'at least' (probability of outcomes of this much or better) and 'at most' (probability of outcomes this low or lower) values. When playing games where there's no downside for higher success scores, you want the 'at least' scores. For games where overshooting an intended score can be dangerous, (I know of only one such game, but it doesn't use dice pools), you should look at the sum of exact probabilities of all scores in an acceptable bracket - that's your chance of getting the desired outcome.
Best Answer
Use a Tree Diagram
Because I am bad at remembering formulas and bad at probabilities in general I like to use a tree diagram. The other answers provide good straight forward solutions you can apply but I hope to show why those methods work.
For Case 4: Rolling 4 d20s and taking the lowest value
If we roll a d20 4 times, the tree will have four levels with two branches at each level: You either roll a 7 or higher, or you don't. At each branch we assign a probability for that individual outcome: in this case, a 6/20 for rolling a 1-6, and a 14/20 for rolling a 7-20. In this diagram I haven't filled out the branches where we roll a 1-6 because they're not relevant for this question.
To calculate the final probability, we need each roll to result in a 7-20. For this, we take the probability of each branch and multiply them together, so we get
$$\frac{14}{20} \cdot \frac{14}{20} \cdot \frac{14}{20} \cdot \frac{14}{20} = \left(\frac{14}{20}\right)^4 \approx 24\% $$
For 15 or higher in this case, we can substitute 14/20 with 6/20 and hence get $$\frac{6}{20} \cdot \frac{6}{20} \cdot \frac{6}{20} \cdot \frac{6}{20} = \left(\frac{6}{20}\right)^4 \approx 0.8\% $$
For case 3: Rolling 3 d20s and taking the lowest
Similarly if we were to look at 3 dice rolls instead, the tree would only have 3 levels, and hence we would only end up multiplying the probability 3 times, so we would get
$$\frac{14}{20} \cdot \frac{14}{20} \cdot \frac{14}{20} = \left(\frac{14}{20}\right)^3 \approx 34\% $$
For cases 1 and 2: taking the highest value
Now if we were interested in finding the probability of getting 7 if we took the highest roll each time we would have to consider the likelihood of getting a 7-20 once anywhere in the tree, not just the branch where we get it 4 times in a row, which would be more complex. Luckily there's a trick for these kinds of situations: Instead of considering the probability of getting at least 1 7-20 in 4 rolls, we can consider the probability of never getting a 7-20 in 4 rolls and our result will be 1 minus that probability.
So to do that we must first find the likelihood of getting a 1-6 4 times in a row, and we would see that it is as above
$$ \frac{6}{20} \cdot \frac{6}{20} \cdot \frac{6}{20} \cdot \frac{6}{20} = \left(\frac{6}{20}\right)^4 \approx 0.8\% $$
Therefore we must take 1 (or 100%) and subtract the probability we found which gives us a 99.2% chance of rolling at least a 7 in 4 rolls. Likewise for 3 rolls we get 97.3%