You need to find two things, the open circuit voltage and the short circuit current. To find the open circuit voltage, remove RL from the diagram and find the voltage where the 12Ω resistor connects to the 8Ω resistor. I get the following sequence of circuits as I simplify.
You should probably check my math. From this I determine that the voltage between the 4Ω and 20Ω resistors is 60 v. That means there is a 12 v drop across the 4 Ω resistor, but that resistor is 20Ω in parallel with 5 Ω, and the 20Ω was 12Ω+8Ω, so there is a 12 v drop across the 12+8Ω, or 7.2 v drop across the 12Ω, thus the open circuit voltage is 72 - 7.2 = 64.8 v.
The next thing we need is the short circuit current. Replace RL with a short to ground. I get the following:
You now need the short circuit current, which is the current through th 12Ω resistor and the current throught the 8Ω resistor, I get 6A through the 12Ω and 4.8 A through the 8Ω.
Rth = Voc/Isc. I get Rth = 6Ω
Vth = Voc.
From this you should be able to derive the Norton equivalent circuit. You had best check my math.