"What is the probability the word contains "H" once?"
This is slightly ambiguous. Do you mean the word i) contains at least one H or ii) contains exactly one H?
i) has a trick. There are are $29^5$ total words. There are $28^5$ words that contain no H's whatsoever (-- each letter can be any one of the remaining 28 letters). So there are $29^5 - 28^5$ words that have one or more Hs.
ii) there is a naive way and a sophisticated way. First the naive way: There are $1*28*28*28*28$ ways to type a five letter word a five letter word starting with H and the rest of the letters are not H. There are $28*1*28*28*28$ ways to type a word where the 2nd letter is H but the rest are not H. There are $28*28*1*28*28$ where the third letter is H and so on.
So in total there are $1*28*28*28*28 + 28*1*28*28*28 + ..... + 28*28*28*28*1 = 5*28^4$.
... or the sophisticated way: The number of ways to type a 5-letter word where a specific letter slot must be H and the remaining 4 letter slots must not be H is $1$ for the dedicated letter slot and $28^4$ for the remaining 4 letters slot (each of which can be any of the 28 remaining letters remaining). That is $28^4$ possible ways. Now there are 5 possible choices for which dedicated letter slot is the H. That means there are $5*28^4$ possible words. (5 choices for which letter is the H-- and 28 choices for each of the remaining four letters. Multiply the choices... $5*28^4$.
====
" 2*2/3*3. I don't see the connection to the 5 in the answer"
If you need to figure out $n$ letter word with $m$ choices of letters the number of words with exactly one ~ is $n*(m-1)^{n-1}$. So the answer with 3 letters and a 3 letter word is $3*(2)^{2} = 12$ [ABB,ABC, ACB,ACC, BAB,BAC, CAB,CAC, BBA,BCA,CBA,CCA].
One 2 corresponds the 28. The other to the 4. The 3 corresponds to the 5.
====
Two post-scripts.
I just realized I gave the answers to "What is the NUMBER OF WAYS" rather than "What is the probability".
Well, Probability of Event = (Number of Ways Event can Happen)/(Number of total things that can happen).
So for i) Number of ways to get at least one H = $29^5 - 28^5$. Number of total things that can happen = $29^5$.
So probability is $(29^5 - 28^5)/29^5 = 1 - (28/29)^5$.
for ii) Number of ways to get exactly one H = $5*28^4$. Number of total things that can happen = $29^5$ so probability is $5*28^4/29^5$.
For your 3 letters and a three letter word: Number of ways to get exactly one A = $3*2^2 = 12$. Total ways to type three letters = $3^3 = 27$.
So probability = $3*2^2/3^3 = 2^2/3^2$. Note: because you had 3 letters total and 3 letter length words one of the 3s cancelled. Which is probably why you didn't see the corespondence.
If you had say a four letter word of ABC with exactly one A then the total ways would be $4*2^3/3^4$. (Do you see why) and no canceling. That you chose the same number of letters and length of words was a misleading coincidence.
2nd postcript as per aarons comment.... well, I have to go cook dinner now. I'll get to it later but... It's to introduce the idea of choicing m out of n options. i.e. how to choice which position the one H is. Or if you had to calculate the probability of a five letter word with exactly TWO Hs--- how to choice which of the 5 positions we can place the 2 Hs....
Think about it. I'll get back to it.
Best Answer
A experiment with real monkeys suggested that the keys pressed are not independent - in that case they repeated letters a lot
Back to your question, it might be sensible to subtract $k-1$ from $n$ as you cannot get your desired string of $k$ characters with in the first $k-1$ attempts.
Ignoring that point, an approximation with large $n$ is $$1 - \left(1 - \frac{1}{k^m}\right)^n \approx 1- e^{-n/k^m} $$
If you want the right hand side to stay constant as $k$ increases by $1$ then you want $\frac{n_1}{(k+1)^m} \approx \frac{n_0}{k^m}$ so you want $\frac{n_1}{n_0} \approx \left(1+\frac1k\right)^m$.
Whether $e^{m/k}$ is a good approximation to that ratio depends on the particular values of $m$ and $k$