To find \$I_3(2)\$ you need to know the voltage at the top node (call it \$V_t\$). Then you know by Ohm's Law that $$I_3(2) = (E_3 - V_t)/Z_3$$
To find \$V_t\$ you can use superposition (assuming \$E_3\$, \$E_1\$, and \$I_g\$ are independent sources). This requires turning off all sources except one and finding the remaining source's contribution to \$V_t\$, then repeating until you have found the contribution of all three sources. \$V_t\$ due to all three sources is then by superposition the sum of the three contributions.
To turn off a voltage source make it a short circuit, and to turn off a current source make it an open circuit.
For example, to find the contribution of \$E_3\$, \$E_1\$ becomes a short and \$I_g\$ becomes an open circuit. That means \$Z_1\$ and \$Z_2\$ are in parallel, and this parallel resistance is in series with \$Z_3\$. \$V_t\$ is then the output of a voltage divider formed by \$Z_3\$ and \$Z_1 \parallel Z_2\$.
\$E_1\$'s contribution is very similar, except \$Z_2\$ and \$Z_3\$ are in parallel and this parallel resistance is in series with \$Z_1\$.
For \$I_g\$'s contribution, with \$E_1\$ and \$E_3\$ off all three impedances are in parallel so \$V_t = I_g \times (Z_1 \parallel Z_2 \parallel Z_3)\$.