The 2 examples you made (the bible and the sequence of the number $\frac{1}{3}$) are similar, but deeply different. In fact, the bible is a finite sequence of characters, whereas the decimal digits of $\frac{1}{3}$ are infinitely many.
So, the probability that the monkey writes exactly the bible during its typing is 1, while the probability that the monkey will write exactly all the digits of $\frac{1}{3}$ is 0.
Digits of $\frac{1}{3}$
In order to write exactly the digits of $\frac{1}{3}$, from a certain digits the monkey must push the key 0 and then the key 3 infinitely many times... up to infinity! This has probability (not rigorous) $\frac{1}{10} * \frac{1}{10} * \dots * \frac{1}{10} = (\frac{1}{10})^\infty = 0$.
Sequence of letters of the bible
In order to write exactly the sequence of letters of the bible, instead, the monkey should guess a finite number of characters.
The first thing we can notice is that the probability that this happens is larger than 0. In fact, assuming that the bible is 1.000.000 letters long, the probability that the bible starts from the first letter of the first page that the monkey writes is something like $p = \frac{1}{26} * \frac{1}{26} * \dots * \frac{1}{26} = (\frac{1}{26})^{1.000.000}$, which is a very small number, but it is larger than 0.
Then, think that the bible could start from the second letter of the first page, so we have another $p$ probability that the monkey will write exactly the bible. And consider that the monkeys has infinitely many letters from which it can start to write the bible, so it has infinitely many chances to succeed in the challenge, each of them having $p$ probability to work.
Note that these infinitely many events (each corresponding to which letter the monkey is starting to write the bible from) are not independent, so the total probability will not be $\infty * p = \infty$, but they are enough to guarantee that the monkey will achieve it anyway, with probability 1.
The rigorous proof that this probability is 1 is not even difficult, but I will not write it here. You can find it on Wikipedia.
The idea of the proof is to estimate the probability that the monkey will not write the bible and eventually you can proof that that probability is 0, meaning that it is almost impossible (but still not impossible) that the monkey doesn't write the bible.
Question: Do these sequences truly run forever?
Answer: Yes. There is a finite probability that the process never ends. Let $E_n$ be the event that the game ends after exactly $2n$ flips. Note that
$$
n\ge 2\implies P(E_n)\le 2^{-n}-2^{-(n+2)}
$$
In order for $E_n$ to occur, the last $n$ flips must match the first $n$ flips, which occurs with probability $2^{-n}$. However, the $2^{-n}$ calculation includes sequences that would have ended after fewer than $2n$ flips. For example, any sequences which looks like this will be counted, even though it would have ended after the first two flips.
$$
0,0,\underline{\;\;\;\;\;\;\text{series of $n-2$ flips}\;\;\;\;\;\;},0,0,\underline{\;\;\;\;\;\;\text{same series of $n-2$ flips}\;\;}
$$
The probability of the above sequence is $2^{n-2}/2^{2n}=2^{-(n+2)}$. There are many other cases which need to be subtracted. However, subtracting only the above case gives an upper bound.
We then apply the union bound; the probability of at least one of the events $E_1,E_2,E_3,\dots$ occurring is at most the sum of their probabilities. This sum is at most
$$
P(\text{process eventually ending})\le \sum_{n=1}^\infty P(E_n)\le \frac12+\sum_{n=2}^\infty (2^{-n}-2^{-(n+2)})=\frac12+\frac12-\frac{1}{8}\le \frac78
$$
In particular, with probability at least $1/8=12.5\%$, the process will continue forever. Above, I used the infinite geometric series formula to compute the infinite sum.
Question: How does this relate to the law of large numbers, which says that infinitely repeated events will certainly eventually occur?
Answer: In general, if you have a series of independent events $E_1,E_2,\dots$ with the same nonzero probability, then you can be certain that these events will occur infinitely often. However, for your problem, the events are dependent. Also, the probabilities are not the same, but are decreasing over time. These two difference are why you get a different result.
It is possible to weaken the assumption about the same nonzero probability. In general, if $E_1,E_2,\dots$ is a sequence of independent events such that $\sum_{n=1}^\infty P(E_n)=+\infty$, then you can conclude that infinitely many of the events will occur. This is the second Borel-Cantelli lemma. For example, if $P(E_n)=1/n$, then you can conclude that there will be infinitely many occurrences.
Best Answer
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:
and also eventually type:
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.