A Series-parallel circuit combination with current as input source
The voltage across the 4-ohm resistor, |V|, is the voltage across the left-side 12-ohm resistance (\$V_a\$) minus the voltage across the 6-ohm resistance, (\$V_b\$). Since you know the voltage across the 8-ohm resistance (72 V), you can determine \$V_a\$ by calculating the equivalent resistance loading the left-side 12-ohm resistance: \$R_{eq}=12||(4+6||12||4)\$. For the symbolic expression, do not develop the terms and keep the || as it is easier to read. Once you have that \$R_{eq}\$ value, you see a resistive divider with the 20-ohm resistance and \$R_{eq}\$. From the 72-V level you have calculated, apply that resistive divider and obtain \$V_a\$. Now that you have \$V_a\$, apply another resistive divider considering the 4-ohm resistance and the parallel combination of 6, 12 and 4 to obtain \$V_b\$. Calculate \$V_a-V_b\$ and you have what you want. With \$V_b\$, you have \$I_1\$ and \$I_2\$.

As a side note, below is a resistive divider involving \$R_1\$ and \$R_2\$ in the left side.


The voltage at node a is equal to \$V_a=V_1\frac{R_2}{R_2+R_1}\$. If you now load \$R_2\$ with \$R_3\$ as shown in the right side, you have an equivalent resistance equal to \$R_{eq}=R_2||R_3\$ and the new division ratio becomes \$V_a=V_1\frac{R_{eq}}{R_{eq}+R_1}\$.