Help finding maximum power transferred to load RL
Your analysis is incorrect. For one, the 6V source is not all across the 3k resistor so the current thru the 3k resistor is not 2 ma.  The easiest way to solve this is to first find the thevenin equivalent of the 6v source and the 3k and 6k resistors.  The equivalent voltage is simply (6/9)*6 or 4 volts.  The equivalent resistance is 3k in parallel with 6k or 2k. So now you have a 4 volt source in series with a 2k resistor.  To find the equivalent resistance of the complete circuit, you short the 4v source and open the current source.  Then you just have two 2k resistors in series.  So the equivalent resistance of the circuit is 4k.  To find the equivalent voltage, note that no current can flow through the 4v source and its 2k equivalent since RL is now an open circuit.  The only voltage is the 2ma current source through the other 2k resistor which yields 4 volts. So the equivalent circuit that RL sees is 8 volts (4 from the 4 volt source and 4 from the current source through the 2k resistor) in series with a 4k resistor. For maximum power, RL must equal 4k.  Half the voltage will be across it or 4 volts. 4 volts across 4k is 4 mw.