how do we find the Norton equivalent circuit current of the both node ,a-b
Addressing your points in order,

1) I wonder if the solution they have is using a transformation of the source. You can convert the 2nd circuit you have to the following one via source transformation. The voltage source in series converts to a current source in parallel with that same resistor. The current source's rating is calculated using V = IR, where R was the resistor that was in series (so R1 in your drawing). Thus, I = 36/6 = 6 A. Then, it becomes that the current in each branch is 1/3 of 6 A because they are balanced. In any case though, the resistors are in parallel now, so that is where they get the 6/6/6 to find the voltage, so (1/6 + 1/6 + 1/6)^-1 = 2 ohms. V = 6 * 2 = 12V. You have the voltage now over that resistor to ground, so you have the current. I think this solution seems complicated however and would do what you did in point 1.

2) I believe the 36/6 value is from the source transformation that we did above. 

3) Ignoring the load resistor is an important part of setting up Norton's equivalent circuit. The important step is that you short the load (R4) and short independent voltage sources. You make independent current sources open as well.

This wiki page might serve as a good refresher.

https://en.wikipedia.org/wiki/Norton%27s_theorem






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