With respect to your first question, because a digital representation of Graham's number is a finite text, your infinite monkey theorem would apply, so a monkey would eventually type out Graham's number. However, if you replace the monkey with the infinite decimal representation of an irrational number, then it depends on the number.
As obvious examples, one "known" irrational number (Champernowne's constant) is the number that consists of a concatentation of all natural numbers:
$$0.123456789101112131415161718192021222324\ldots$$
This clearly contains Graham's number. Another "known" irrational number is:
$$0.10110111011110111110\ldots$$
and this clearly doesn't contain Graham's number (since, for one thing, Graham's number ends in a seven).
It is unknown whether $\pi$ or $e$ or $\sqrt{2}$ contain Graham's number. It's also unknown whether any of these are "normal numbers", which is a precise way of describing the sort of "uniform distribution of digits" that you mention in your question. If one of these numbers was normal, then it would behave a lot like a monkey and therefore would contain Graham's number. But, if it was non-normal, it still might contain Graham's number, or it might not. Anyway, we just don't know.
For your second question, the answer is that this assertion is true, and it more or less follows from the infinite monkey theorem as you've stated it, provided you accept that, if you arrange your countably infinite monkeys in a row, then the string of digits they produce each second has the same distribution as the countably infinite string produced by a single monkey over infinite time.
If you accept that the output of infinite monkeys each second is equivalent to the infinite output of one monkey, then you just have to observe that, by the infinite monkey theorem, a single monkey would eventually type:
Here is the first copy of the entire works of Shakespeare: [entire works]
and also eventually type:
Here is the second copy of the entire works of Shakespeare: [entire works]
and so on. Each occurrence in this countably infinite set of copies would represent a separate copy of the entire works of Shakespeare which -- getting back to our equivalent row of monkeys -- would represent a separate subset of monkeys who typed the entire works of Shakespeare within the same second. This argument applies every second, so each second, a countably infinite set of subsets of monkeys each generate the entire works of Shakespeare.
For your third question, yes, each second every one of the 88 notes are played by a countably infinite collection of monkeys. There is no second where a note is unplayed. In other words, even though many unlikely things occur countably infinite times during the recital, there are some things (like a note being missed) that never occur.
This more or less follows from the same argument I gave above for your second question. The sequence consisting of all 88 piano notes in order is finite, so it must be played at least once (in fact, a countably infinite number of times) by a single monkey over infinite time. By the equivalence to a countably infinite row of monkeys playing within a single second, all 88 keys must be played by a subset (in fact, a countably infinite number of subsets) of monkeys in a second. This argument applies every second, so every second, all 88 keys are played by each of a countably infinite number of subsets of monkeys.
Best Answer
No, $X_n$ is the chance that in $n$ monkey-blocks there will not be a 'banana' that we recognize. If we have $100$ billion monkey-blocks, either from $1$ monkey typing $600$ billion characters or $100$ billion monkeys typing $6$ characters each the chance that there is no recognized 'banana' is $0.0017$. That means the chance we do have at least one recognized 'banana' is about $1-0.0017=99.83\%$