You do have events $H_1$ and $H_2$, namely, they are the events
you labeled $A$ and $B$:
\begin{align}
H_1 & = \{ HH, HT \}, \\
H_2 & = \{ HH, TH \}. \\
\end{align}
In general, if $HH$, $HT$, $TH$, and $TT$ are simply the labels of events
in some arbitrary probability space $\Omega$,
these events might or might not be independent.
The sum of their probabilities merely needs to be $1$.
We do usually suppose, however, that a good model of two successive tosses
of the same coin would be a probability space for which
$P(H_1) = P(H_2)$ and for which $H_1$ and $H_2$ are independent.
The fact that $P(HH) = P(H_1 \cap H_2) = P(H_1)P(H_2) = (P(H_1))^2$ follows immediately.
Let $P(H_1) = p$ as a more convenient way of writing $P(H_1)$,
and then $P(HH) = p^2$, $P(HT) = p(1-p)$, and so forth by simple algebra.
The binomial distribution PMF isn't really describing the same thing - it looks for the possibility of getting a specific exact number of tails. If that's really what you're looking for (the probability that 3 are tails and the rest are heads), the replacement for nCk is n-k+1, or 3, because there are 3 places where there could be 3 consecutive tails (the beginning, the middle 3, or the last 3). Following this should get you 3/32.
On the other hand, if you're looking for the probability of there just being 3 consecutive tails (so that 5 tails in a row would also count), you're going to have a lot of difficulties with overcounting. The easiest method in this case is really just to count the possibilities (there are 8 - ttttt,tttth,tttht,ttthh,thttt,htttt,httth,hhttt). Divide this by the total number of possibilities (there are 32 ways to flip a coin 5 times) to get the overall probability of 1/4.
If you really want to do the second part in a slightly fancier way, you have to avoid counting sequences like ttttt multiple times. The way to do this is to look for an h followed by 3 t's - if you find httt, there can be no previous sequence of 3 t's, and you avoid overcounting. Therefore, you are looking for ttt??,httt?, and ?httt, which have probabilities of 1/8,1/16,and 1/16 respectively, summing to 1/4.
Note that this method won't work for significantly longer sequences, as it's possible to have 3 consecutive t's, then httt. More advanced methods are required.
TL;DR depending on what you mean, the answer is probably 1/4, but the method you described isn't particularly useful
Best Answer
$P$(more than 50 heads) = $\sum_{n=51}^{90} P$ (exactly n heads) = $\sum_{n=51}^{90} {90 \choose n} (1/2)^{90} = (1/2)^{90}\sum_{n=51}^{90} {90 \choose n} .$ You are right. If $X$ denotes the number of heads achieved in $90$ tosses then $X \sim Binomial(n,1/2)$, assuming the coins are fair.