Mesh analysis - current source in one mesh?
The currents split up as shown in the following schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

You could easily combine \$R_1\$ and \$R_2\$ into a single effective value and work out the voltage at their shared top node... and from that work out their currents. But you can also just as well see that \$R_1\$ is 3 times less conductive than \$R_2\$, so this means that if one part of the current flows through \$R_1\$ then it must be that three more parts of the current must flow through \$R_2\$. From this argument, you should consider dividing the total current of 12 A into four parts of 3 A each. And thus, \$i_1=3\:\textrm{A}\$ and \$i_2=9\:\textrm{A}\$.

Similar logic then applies to the other two resistors. Here, it's not hard to imagine a total of 7 parts of the -2 A net total current in \$R_3\$ and \$R_4\$, where now three parts go through \$R_3\$ and four parts through \$R_4\$. Each part is \$-\tfrac{2}{7}\:\textrm{A}\$, so this means \$i_3=-\tfrac{6}{7}\:\textrm{A}\$ and \$i_4=-\tfrac{8}{7}\:\textrm{A}\$.

Looking at the bottom node, we know the following must be true:

$$ -8\:\textrm{A} - 2\:\textrm{A} + 3\:\textrm{A}+9\:\textrm{A}+-\tfrac{6}{7}\:\textrm{A}+-\tfrac{8}{7}\:\textrm{A}=0$$

And it is.