Trying to compute the power that the operational amplifier provides to a load
Lets say VCC = 5V and VEE = -5V, the maximum amplitude for the output
  signal will be 5V=VCC ?


Not exactly. The output transistors are in Emitter Follower configuration, so the output (at the Emitter) will be ~0.7 V less than the voltage on the Base. The current flowing through RC3 will raise the voltage on T4's Base, while RE3 and T3 (in saturation) will lower the voltage on T5's Base. The result is peak to peak output voltage will be less than the rail-to-rail voltage by at least 1.4 V.   


  what is the point of having R3 in there? And what would the value for
  R3 has to be taking into regard what i said about R1, R2 and the gain.
  It is said to be ~R2//R1 but i dont know why.


R3 balances the offset voltage on the inverting input caused by input bias current flowing through R1 and R2. It is made the same value as R1||R2 to create the same offset voltage on the non-inverting input.


  to compute the input impedance of that circuit (op-amp with negative
  feedback) what should i do? Can you guys give me any tips?


With negative feedback The inverting input is driven to the same voltage as the non-inverting input, ie. ground, and so is called a virtual ground. Imagine R1 is connected to ground and the input impedance is obvious.


  Now this op-amp with negative feedback is said to have a gain of 10. I
  need to choose the dimensions of R1 and R2 to make it right.


If the open-loop gain was infinite then R2 would need to be 10 x R1 ie. 50k. With an open-loop gain of only 130 it must have less feedback to produce a closed-loop gain of 10, so R2 should be higher than 50k. So there must be a mistake somewhere in your calculations.   


  Last question, i need to find the current source (IF1) output
  impedance. I believe i need to do the small signal and try to obtain
  it from there?


Hard to determine its exact value without knowing the characteristics of T6, but (when inputs are inside the common mode voltage range) it should be close enough to 'infinite' for most calculations.