Average value of an observable I got confused by the concept of the average value of an observable.
I know that when we measure a physical quantity $A$ in a specific state described by $\psi_1$,  we only get the eigenvalue $a$ of the operator $\hat{A}$. 
So theoretically the value of the physical quantity $A$ that we are measuring is equal to the eigenvalue of $\hat{A}\psi_1=a\psi_1$.
Then there is a formula to find the average value $\langle A\rangle$. And this is also an expectation value.

Does it mean that normally $\langle A\rangle=a$?

Then I get confused by the world ''average'' in ''average value $\langle A\rangle$''. Is this the average value of the physical quantity $A$ in all its the states?

And finally, are we assuming here that my system has only one system described by $\psi$? For example there is only one particle. Is it possible that I have three systems (three particles) described by different eigenfunction $\psi$ and eigenvalue $a$?
I am confusing everything.
 A: You are confusing a lot all right : )
Start with getting this equation $  \hat{A}|\psi_1 \rangle =a |\psi_1\rangle$ nailed  down first. 
You apply a Hermitian  operator, $\hat{A}$ to a ket $|\psi_1 \rangle$ and it  gives you an appropriate  eigenvalue $a $.
So the  $ \hat{X}$ operator gives you the position observable $x$, likewise applying $ \hat{P}$ gives you the real observable momentum $p $,  associated with  $|\psi_1 \rangle$.
I am assuming you know what using a Hermitian operator implies.

Does it mean that normally $<A>=a$? 

You should read this website article later: Average Value for a fuller description but a summary for now is:
No, $<A>$ is not equal to $a$, in general. 
$<A>$ is a prediction of the average value of repeated measurements, whereas  $  \hat{A}|\psi_1 \rangle =a |\psi_1\rangle$ is the value of 1 (just 1) single measurement.
For example, say you have a wave function that you know describes the ground state of a particle $m $. 
Now you want to find the average momentum $$<P> = \langle \psi_1 | \hat P|\psi_1\rangle$$
For our purposes, expectation value and average value are the same idea.
So when you work out $<P> = \langle \psi_1 | \hat P|\psi_1\rangle$ in the 1D ground state simple box potential, you are going to find that it's equal to 0.
That's because the average momentum of the particle in these conditions is zero. The particle has the same momentum in two opposite directions, so it's average value is zero. 

Then I get confused by the world ''average'' in ''average value $\langle A\rangle$''. Is this the average value of the physical quantity $A$ in all its the states?

No....   It's the average value in one state, where you sandwich the A between, for example the functions for the ground state.
If you then want to find the average value for the first excited state, you sandwich the A between the functions that describe the first excited state.
You then keep doing this for any  energy state that that you know  the functions  for.

Image Source: Energy Levels

And finally, are we assuming here that my system has only one system described by $\psi$? For example there is only one particle. 
Yes, there is just 1 particle. No offence, but I would absolutely,  definitely,  completely stick to learning about  one particle state for the moment.   

Is it possible that I have three systems (three particles) described by different eigenfunction $\psi$ and eigenvalue $a$?  

No.  I would not worry at this time about  3 particle systems, that is  normally further down the course, that is weeks or months way if you are just starting off. 
A: The average value of some observable is defined specifically as the average value of measurements on an ensemble of states. Think of it like a shelf and in that shelf you have 1,000,000 jars. In each jar you have the same state $\Psi$. Then you have 1,000,000 graduate students who each measure, say the $x$ position of all of these identical states. The average of all of these values will be your expectation value, $⟨x⟩$. So the expectation value is the average value of measurement on the same state, which is the crucial part.
A good example of this would be with a spin $\frac{1}{2}$ particle such as an electron. Say we have an electron in the $|+x⟩ = \frac{1}{\sqrt{2}} |+z⟩ + \frac{1}{\sqrt{2}} |-z⟩$ state, and we want to know the expectation value of the observable $S_z$ in this state. Well, you can compute it with the formula 
$$ ⟨S_z⟩ = \sum_{+,-} |c_n|^2 S_n,$$ or you can think about it conceptually. Again, you have 1,000,000 jars with this state in them. You get 1,000,000 PhD students now to make a measurement on the states, of the spin in the $z$ direction. You would expect that half of them be spin up ($+\frac{\hbar}{2}$) and have of them spin down ($-\frac{\hbar}{2}$), which we can also see from the coefficients of the $|+x⟩$ state. Thus, the expectation value $⟨S_z⟩ = 0$ in this case. I hope this clarifies your doubts.
A: A quick clarification on average, mean and expectation value.
Here are two arrays of dice throws obtained with the following random generator:
$$\{x_1\}=(2,2,1,6,5,2,4,6,6,5)$$
$$\{x_2\}=(3,4,4,4,5,2,6,2,2,4,4,5,5,3,2,2,5,2,1,1)$$
The sample average (sample mean) of both is:
$$\bar{x}_1=3.9$$
$$\bar{x}_2=3.25$$
Of course we know that the true mean (or population mean) is given by:
$$\langle\mu\rangle=\frac{1+2+3+4+5+6}{6}=3.5$$
This is what's known in quantum physics as the expectation value. But it's clear that this term is, in a sense at least, very much a misnomer: does anyone really expect to throw a $3.5$? Of course not: the probability of throwing a $3.5$ is $0$!
Similarly, the wave function of a particle in a 1D box of length $a$ is:
$$\psi_n(x)=\sqrt{\frac2a}\sin\Big(\frac{n\pi x}{a}\Big)\:\text{for } n=1,2,3,...$$
The expectation value of the position $\langle x\rangle$ can easily be calculated to be:
$$\langle x\rangle=\frac{a}{2}$$
But it can also be shown that the probability of finding the particle at $x=a/2$ is in fact $0$!
