Electronic – KVL loop current direction convention–should be same in all loops

I have no idea what setup I should use for the two meshes with the ideal current source.

Here's a hint: the current source provides the following equation:

\$ 1A = I_C - I_B \$

This is analogous to having a floating voltage source when doing node voltage analysis. There, you must form a supernode enclosing the voltage source and add the equation relating the voltage between the two nodes enclosed to complete the solution.

Here, you must form a supermesh and add the equation I've given above.

Strictly speaking, it's not the sign of the resistors that can be changed, it's the sign (direction) of the current \$I_x\$. Resistors always have positive resistance but you can choose the current in either direction as long as you are consistent.

Suppose you chose \$I_x\$ in the opposite direction. Starting from node \$f\$ and moving counterclockwise, you have a positive voltage across the \$6\Omega\$ resistor, another positive voltage across the \$3\Omega\$ resistor, and a positive voltage across the 12V supply. This gives:

\$I_{x}(6\Omega + 3\Omega) + 12 = 0\$

which results in \$I_{x} = -4/3\$A. The sign is different, indicating that the current actually flows in the opposite direction (i.e. in the original direction shown in the image).

The reason your 4A answer is wrong is that you weren't consistent with the \$I_x\$ direction for the two resistors -- you had \$I_x\$ flowing in the opposite direction through the \$3\Omega\$ resistor.