Why does Binomial distribution fits Normal distribution I use Statgraphics for working with statistics.
I generated 3 Binomial samples:
1) N=100, p=0.5
2) N=100, p=0.01
3) N=100, p=0.99
First sample looks like Normal sample:

And it passes Kolmogorov-Smirnov test:

So I have theoretical question from probability theory: why first sample acts like Normal and why other two don't?
 A: Any sum of a large number of identically distributed variables starts to look like a normal distribution; it's just that your first sample reached this approximation more quickly because it started out more "evenly". It's still discrete, though: the "convergence" that we are looking at does not refer to a convergence of density functions.
A: Let $X_1, X_2, \dots$ be a sequence of independent, identically distributed random variables with common mean $\mu$ and variance $\sigma^2$. Then the sum $T_n = \sum_{i=1}^n X_i$ of the first $n$ of the $X_i$ has $E(T_n) = n\mu$ and
$Var(T_n) = n\sigma^2$. Thus $Z_n = \frac{T_n - n \mu}{\sqrt{n}\sigma}$ has $E(Z_n) = 1$ and $Var(Z_n) = 1.$
The Central Limit Theorem (CLT), says that the sequence $Z_n$ converges
in distribution to standard normal. That is,
 $P(Z_n \le z) \rightarrow \Phi(z)$, for any real $z$, where
$\Phi$ is the standard normal CDF. (A similar statement can be made
for averages $\bar X_n = (1/n)\sum_{i=1}^n X_i$ and 
$Z_n = \frac{\bar X_n - \mu}{\sigma/\sqrt{n}}.$)
Whenever you want to put a limit theorem to practical use (such as approximation),
your first question should be "how fast is the convergence?"
Infinity itself is a 'long way away' Roughly speaking, the
convergence in the CLT is more rapid for symmetrical random variables $X_i$
than for skewed ones. Thus, the sum of only ten independent
$X_i \sim Unif(0,1)$ (symmetrical) is very nearly normal, while the sum of 25 independent random variables $X_i \sim Exp(1)$ (markedly skewed) is clearly not so
well approximated by a normal distribution. The diagram below illustrates this with 10,000
such sums of each type of random variable. The appropriate normal
density is shown with each histogram.

The speed of convergence is quite fast in some instances and
rather slow in others. A binomial random
variable $X \sim Binom(n, \theta)$ is the sum of $n$ independent Bernoulli random variables with success probability $\theta.$
So, for sufficiently large $n$ a Binomial random variable will have
a distribution that is 'approximately' normal. In the
binomial case, the convergence is much faster if $\theta \approx 1/2$
(symmetrical) than if $\theta$ is near 0 or 1.
You have already discovered that $Binom(100, .5)$ is closer
to normal than $Binom(100, .01).$ 
Sometimes a 'rule of thumb'
is used. It says that the approximation of normal to binomial
is 'reasonably' good if $n\theta$ and $n(1-\theta)$ both exceed 5.
Few 'rules of thumb' are always accurate. There are better
rules for normal/binomial fit, but this one is often quoted, 
probably because it is usually pretty good and it is easy to remember.
A: One reason is Stirling's formula.

Instead of central limit theorem, you can use De Moivre–Laplace theorem. See the second of two proofs on Wikipedia which makes use of Stirling's formula. De Moivre-Laplace theorem stated on Wikipedia contains this approximation:

Therefore, the first step of the proof is rewriting $\binom{n}{k}$ as factorials and approximating with Stirling's formula

