Action of the state I have the following question:
let $A$ be a C*-algebra and let $a$ be a self adjoint element of $A$. Is it true that
for any state $f$ acting on $A$ $$f(a) \in \mathbb{R}.$$
Let me remind that a state is a positive linear functional of norm $1$.
I think it is due to the fact that every state has to satisty, 
$f(x^*)=\overline{f(x)}$, for all $x \in A$.
Then we easily obtain
$f(a)=f(a^*) = \overline{f(a)}$, thus $f(a) \in \mathbb{R}$, but I don't know how to show that it has this *-property.

Suppose that $a$ is a self-adjoint element in the C$^*$-algebra $A$. Then, by applying the continuous functional calculus, we can write $a$ as the difference of two positive elements $a=a_+ - a_-$ such that $a_+a_-=a_-a_+=0$. See, for example, Proposition VIII.3.4 in Conway's A Course in Functional Analysis, or (*) below.
Once we have this fact in hand it is easy to show the desired property. As $f$ is positive, $f(a_+)$ and $f(a_-)$ are positive (and thus real) and so $f(a)=f(a_+)-f(a_-)$ is real.
It is also now easy to show the self-adjointness property that you mentioned: each $a \in A$ (now not necessarily self-adjoint) can be written as $a=x+iy$, where $x$ and $y$ are self-adjoint. We can take $x=\frac{1}{2}(a+a^*)$ and $y=\frac{-i}{2}(a-a^*)$. Then $$f(a)=f(x+iy)=f(x)+if(y)$$ and $$f(a^*)=f(x-iy)=f(x)-if(y).$$
As $f(x)$ and $f(y)$ are real, we have $f(a^*)=\overline{f(a)}$.
(*) How do we show that $a=a_+-a_-$ where $a_+a_-=a_-a_+=0$ and $a_+$ and $a_-$ are positive?  As $a$ is self-adjoint, its spectrum is a subset of the real line and we also know that we can apply the continuous functional calculus. If $g(t)=\max (t, 0)$ and $h(t)=\min(t,0)$, then $a_+=g(a)$ and $a_-=h(a)$ are the elements we need. Why does this work? Think about splitting a continuous function on $\mathbb R$ into its positive and negative parts.
