Hi, Could you help me how can I resolve this exercises. I'm not really good in this subject , but I'll hope you can explain me, I don't want the answer I only want your help please .

http://k42.kn3.net/C13A22CC3.jpg

 

I will help you with 2nd problem.
Fisrt of all, mark with a-b letters the 3K resistor ends, then remove it from the circuit.
Let´s find out the voltage between a-b points, this is Thevenin´s voltage.
Using Superposition Theorem, remove the current source. Using Source Transformation find the equivalent current source and put 4K resistor in paralell with it. Notice that 4k resistors are in parallel, find the equivalent resistor. Now, apply a current divisor and find the current in 2k-6k branch. Use Ohm´s law multipling this current by 6k resistor.
This is the voltage source contribution to Thevenin´s voltage, say Vth´. In a similar way find out the current source contribution, say Vth´´. Then Vth=Vth´+Vth´´.
Now, we need Thevenin´s resistor. Applying Thevenin´s Theorem short the voltage source and remove the current source from the circuit. Then find the equivalent resistor that points a-b "sees". This is Thevenin´s resistor.
Now, connect the original 3k resistor in series with Thevenin´s equivalent circuit and use a voltage divider. You have encountered Vo.
To get maximum power transfer in the equivalent circuit, the load resistor must be equal to Thevenin´s resistor. You can calculate the power in this way (there are several ways): Pmax=Vth^2/(4Rth).
That´s all.