Superposition Theorem/Nodal Analysis
When you "turn off" a current source, you are essentially inserting a current source of 0A. This is an open circuit, as the open circuit works as an infinitely large resistance, therefore no current is going to flow there. You can simply ignore those wires. The fact they left them in is probably meant to make it visually easier to understand.

In figure (c), there are indeed 3 nodes: one in the middle (v1, common node of the two 4\$\Omega\$ and the 3\$\Omega\$ resistor), one on "top" of the current source (v2), and the one connected to the "bottom" of the source. They decided to make it the reference node, valued 0V for easier calculations. They could have chosen any other node for this, but generally you tend to choose the one on the bottom of your schematic. In most cases, this provides you the easiest solution.

Regarding your question on how they solved (c): From the top of your source, there is 3A flowing out. That has two ways to go from there: through the 8\$\Omega\$ resistor or through the other branch. The sum of these two, obviously, is 3A as well. 
The solution ignores the fact that the second branch also divides later on, because the current flowing through the rightmost 4\$\Omega\$ resistor equals to the sum of currents flowing through the lefmost 4\$\Omega\$ resistor and the 3\$\Omega\$ one. If you look carefully, the voltage drop v2-v1 is through the rightmost, not the leftmost 4\$\Omega\$ resistor.

And lastly, you can choose whichever type of analysis you want, with a little practice you're going to feel which one is easier in a given problem.