For the circuit shown in question 8, we can find a path that includes the one resistor of interest and batteries with no other resistors. The path is identified in green in the image below and I'm going to guess that the current is flowing down in the resistor.

If we apply Kirchhoff's loop rule around this path, starting at the lower left corner and moving in a CW direction, we obtain the following equation. (Remember all the batteries have a voltage of 4.0 V and the resistor has a resistance of 4 Ohms.

If we solve the above equation for I, we get I = 2.0A.