$P$(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.
You are correct that the probability of 0 heads is equal to the probability of 4 heads -- there is one way to get each, and each coin flip has equal probability so the probability for each case is
$$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16}$$
The probability of getting 1, 2, or 3 heads is different. For example, there are four ways to get 1 heads:
HTTT
THTT
TTHT
TTTH
Each of these cases has probability $1/16$ as before, but there are 4 of them so the total probability of getting 1 heads is $4/16$.
It is a similar procedure to count the number of ways to get 2 or 3 heads. More generally, the number of ways to get $x$ heads out of $n$ coin tosses is given by the binomial coefficient:
$$n \choose x$$
To generalize even further, this coin tossing experiment follows the binomial distribution. If the probability of tossing a heads is $p$ then the PMF is given by
$$P(X = x) = {n \choose x} p^{x}(1-p)^{n-x}$$
In this case, for a fair coin $p = 1/2$ so the distribution simplifies a bit.
Best Answer
Let $X_1$ be the number of coin tosses until the first head, $X_2$ be the number of coin tosses after that until the second head, and so on. We want $E[X]$ where $X = X_1 + X_2 + \dots + X_k$. By linearity of expectation, we have $E[X] = E[X_1] + E[X_2] + \dots + E[X_k]$.
For any $i$, the number $E[X_i]$, the expected number of coin tosses until a head appears, is $\dfrac{1}{p}$. You can see this by calculating the mean of the geometric distribution, or by noticing that $E[X_i] = 1 + (1-p)E[X_i]$, etc.
All this gives $E[X] = \dfrac1p + \dfrac1p + \dots + \dfrac1p = \dfrac{k}p$.