I have found that I_1=3.81A and I_3=3.09A and I_2=0.72A and R_th=5.45Ω. What do I need to do to find V_th?
so I need to calculate this loop: But it is an open circuit, does the battery give voltage? does the 4\Omega resistor drop voltage?
Yes, a battery has (gives) voltage, even when it is an open circuit. The resistor will drop voltage only if there's current through it, i.e. if there's a load connected between A and B that draws current.
This diagram has I2 as the current flowing downward in the 2 Ω resistor. But in your original diagram I2 was defined as the current flowing to the right through the adjacent 4 Ω resistor (the one to the left). You need to be consistent in your use of variables. A battery is a voltage source and a voltage source maintains a specified voltage across it regardless of the current through it -- including a current of zero. A resistor drops voltage across it in direct proportion to the current flowing through it -- if there is no current through it, then there is no voltage across it.
Now, if the reason that you have I2 flowing downward in this diagram is because there is no current in the rightmost 4 Ω resistor and therefore I2 must also be the current flowing in the 2 Ω resistor, then that is correct and reasonable.