Op-amp voltage boost circuit : how does it work?
About the gain being stated as \$A_V=3\$, the complete relevant text is 


  R3 and R4 form an output voltage gain stage whose gain, \$A_V=3\$, is reduced to unity at high frequencies by C1 to maintain stability.


What this is saying is that R3 and R4 form a voltage divider so that 

$$v_o = \frac{v_{out}}{3}$$

where \$v_{o}\$ is the voltage at the output of the op-amp IC.

Or, turned around, 

$$v_{out} = 3 v_o.$$

This works because the negative feedback around the op-amp will cause it to push or pull current from its output pin to make it work.

The overall gain of the circuit is 33, as you calculated.


  , it is stated that the gain of the whole circuit is 33 while the gain of the output is 3 . So you can have 2 gains in an op-amp ?


No, the "stage" formed by R3 and R4, with gain 3 doesn't really involve the op-amp. 

But even within the op-amp integrated circuit itself, of course every stage in the design can have a different gain value. 


  So one is the gain of the input to the Vout? and the other is the gain of the op-amp output to the actual Vout?


33 is the gain from \$v_{in}\$ to \$v_{out}\$

3 is the gain from \$v_o\$ to \$v_{out}\$. I think Tim's comment does a better job explaining it than I could:


  
    If the voltage on the op-amp output is not equal to 1/3 of the output voltage (ignoring C1), then current will flow in it's positive or negative power pin. That forms a negative feedback loop composed of the four external capacitors and the op-amps (internal) output stage.