analysis
I need to prove that exists a indefinitely differentiable function $F$ in $\mathbb{R}$ such that $F(x)=0$ if $x\leq a$, $F(x)=1$ if $x\geq b$ and $F$ is strictly increasing in $[a,b]$.
To simplify, I write $f:=\left.F\right|_{[a,b]}$.
My first try use the exponential function. The function $$f(x)=e^{\frac{x-b}{x-a}}$$ solves the problem for $x\rightarrow a^{+}$. $f(b)=e^{0}=1$, but $F$ will not be differentiable at $b$.
Any tips?
A necessary and sufficient condition is that $f^\prime\ge0$ everywhere and $f^\prime > 0$ on a dense subset of $\mathbb{R}$. The condition stated in the question is sufficient, but not necessary. Consider the following differentiable and strictly increasing function for which $f^\prime > 0$ does not hold almost everywhere. Let $C$ be the fat Cantor set (which is closed with positive measure, but does not contain any nontrivial open intervals). Let $g(x)$ be the distance from any point $x$ to $C$. Then, $g$ is continuous and vanishes precisely on $C$. The integral $$ f(x)=\int_0^xg(y)\,dy $$ is strictly increasing and continuously differentiable, but $f^\prime=0$ on $C$.
We can show that $f^\prime\ge0$ everywhere and $f^\prime > 0$ on a dense set is indeed necessary and sufficient for $f$ to be strictly increasing (see also Olivier Bégassat's comments to the question).
Sufficiency: If $f^\prime\ge0$ then the mean value theorem implies that $f$ is increasing. If it was not strictly increasing then we would have $f(a)=f(b)$ for some $a < b$ implying that $f^\prime=0$ on $[a,b]$, contradicting the condition that $f^\prime > 0$ on a dense set. This proves sufficiency of the conditions.
Necessity: In the other direction, suppose that $f$ is strictly increasing. Then, $f^\prime\ge0$ follows directly from the definition of the derivative. Also, $f(a) < f(b)$ for any $a < b$. The mean value theorem implies that $f^\prime(x) > 0$ for some $x\in[a,b]$. So, $f^\prime > 0$ on a dense set.
Aside. Your item 3 sets an unattainable goal. If a continuous function $f$ satisfies $f'=0$ for all but countably many points of $\mathbb R$, then $f$ is a constant function. See Set of zeroes of the derivative of a pathological function.
Here is a more explicit construction than in the post link in comments. Fix $r\in (0,1/2)$. Let $f_0(x)=x$. Inductively define $f_{n+1}$ so that
Here is a piece of this function with $r=1/4$:
and for clarity, here is the family $f_0$,$f_1$,... shown together. All the construction does is replace each line segment of previous step with two; sort of like von Koch snowflake, but less pointy. (It's a monotonic version of the blancmange curve).
To check that the properties hold, you can proceed as follows:
(It may be more enjoyable to prove part 3 using the Law of Large Numbers from probability.)
Best Answer
Define
$$f(x)= \begin{cases} 0, \;\textrm{if}\;x\leq a\;\textrm{or}\; x\geq b, \\ e^{\frac{b-a}{(x-a)(x-b)}}\; \textrm{if}\; a<x<b. \end{cases} $$ Now, define the constant $$c=\int_{a}^{b}f(x)dx=\int_{a}^{b}e^{\frac{b-a}{(x-a)(x-b)}}dx.$$
So, we can take $$F(x)=\frac{1}{c}\int_{-\infty}^{x}f(x)dx.$$
For all $x\leq a$, $F(x)=0$ and, for all $x\geq b$, $F(x)=1.$ $f$ is non-negative, so $F$ is striclty increasing in $[a,b]$. To finish, $F$ is indefinitely differentiable.