KCL for this circuit?
Node V2 is handled the same as node V1: The sum of the currents into the node is zero.  The current in your schematic labeled \$I_d\$ is 4A, due to the independent current source.  So the equation for node V2 is:
$$
\frac{V_1-V_2}{1\Omega} - \frac{V_2}{1\Omega} - 4A = 0
$$
Using this equation and your equation for node V1, you can solve for voltages \$V_1\$ and \$V_2\$.  The voltage at A is \$V_2\$; it is neither 0 nor 8V.