Circuit analysis current?

When i1 splits just above the 12Ω, it sees 12Ω both ways and therefore half goes down into the 12Ω and the other half goes right into the 8Ω+4Ω. Thus V2 = 4*io = 4*i1/2 = 2i1

V1 = 2Ω*i1 * 12Ω*i1/2 = 8i1 SO V2/V1 = 2i1/8i1 = 1/4 and V2 = V1/4

The equations from KVL leads to:

I_1 + ((V_2 - V_1)/8) = (V_1)/12;

((V_2 - V_1)/8) + ((V_2)/4) = 0;

(V_2)/4 = (I_0);

This leads to:

(4(I_0) - V_1)/8 = -I_0

I_1 + ((4(I_0) - V_1)/8) = (V_1)/12:

Solve for V_1 for both equations:

V_1 = 12(I_0);

V_1 = (12/5)(2(I_1) + (I_0));

set them equal and solve for I_0;

12(I_0) = (12/5)(2(I_1) + (I_0));

5*I_0 = 2(I_1) + (I_0);

4(I_0) = 2(I_1);

I_0 = (I_1)/2

This looks suspiciously like homework. You should start by figuring out the equivalent resistance. Then work backwards once you know the total current which is going through the 2 ohm resister. Then you know the voltage drop across that resistor. And you can continue from there.