To find the equivalent resistance between the two points, first note that the 12Ω and 6Ω are in parallel with one another, and so you can combine them as follows:

12Ω//6Ω = 1 / (1/12 + 1/6) = 4Ω

also notice that the 4Ω and 8Ω are also in parallel and can be combined as well:

4Ω//8Ω = 1 / (1/4 + 1/8) = 8/3 Ω

Now you have simplified the circuit to just 3 resistors in a series, and so they can all be combined by simply adding them all together.

R = 4Ω + 5Ω + 8/3 Ω = 11.6666 Ω

Total current can be found through the use of Ohms law:

I = V/R = 36.2 V / 11.6666 Ω = 3.1 A

Current through each combined resistor should be this amount, however, since some are in parallel their combined current will be 3.1 A.

For example the 12 Ω resistor will have current =

total current * (resistor in parallel / both resistors added)=

3.1 A * (6 / (12 + 6)) = 1.03333 A

similarly the 6 Ω resistor will have current =

3.1 A * (12 / (12 + 6)) = 2.0666 A

Notice how the total currents add to 3.1 A? And also note that current flows through path of least resistance, which explains why the 6Ω resistor has more current.

The same steps can be repeated to find the current in the other parallel resistors.

The current in the 5Ω resistor will be just 3.1 A , because there are no circuits in parallel with it.