Resistors in series: R= R_1+R_2+R_3+.....
Resistors in parallel: 1/R= 1/R_1+1/R_2+1/R_3+.....
Start by combining the resistances so that we can work out the current flowing in the various paths.
The 8Omega resistor is in parallel with 14Omega (3+5+6) so the combination (let's call it R_a) is
1/R=(1/8 +1/14) = 11/28
R_a = 28/11 " "(=2.5454 Omega)
R_a is in series with 4Omega and the combination is in parallel with 10Omega, so
1/R_b = (1/10 + 1/( 4+28/11)) = 0.1 +1/(72/11) = 0.1+11/72=0.2528
R_b = 3.9560 Omega
R_b is in series with 2Omega so
R_(Total) = 2 + 3.9560 = 5.9560 Omega
I = V/R = (12)/(5.9560) =2.0148A (total current flowing from battery)
This current flowing through the 2Omega resistor divides into 2 paths, the 10Omega resistor and R_b
It is possible to proportion the currents through R_b and then R_a, but easier to subtract the voltage drops across the 2Omega and 4Omega resistors.
V_(R_a) = 12- (2*2.0148)= 7.9705V
so current through the 4Omega resistor
= (7.9705/(4+R_a)) = 7.9705/(4+(28/11)) = 1.2177A
so V_(4Omega)= 4* 1.2177 = 4.8708V
Voltage across 8Omega is 7.9705 - 4.8708 = 3.0997V
So current through the 8Omega resistor is 3.0997/8 = 0.387A
