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Here's the solution to the circuit problem:

a) Calculate the potential difference across the 4.0 $\Omega$ resistor.

Step 1: Identify the given values for the 4.0 $\Omega$ resistor. The current flowing through the 4.0 $\Omega$ resistor is $I = 1.5 \, \text{A}$. The resistance is $R = 4.0 \, \Omega$.

Step 2: Apply Ohm's Law ($V = IR$) to find the potential difference. $V = I \times R$ $V = 1.5 \, A \times 4.0 \, \Omega$ $V = 6.0 \, V$ The potential difference across the 4.0 $\Omega$ resistor is $\boxed{6.0 V}$.

b) Calculate the resistance of resistor R2.

Step 1: Determine the potential difference across R2. Resistors in a parallel branch have the same potential difference across them. Since R2 is in parallel with the 4.0 $\Omega$ resistor, the potential difference across R2 is the same as calculated in part (a). $V_{R2} = 6.0 \, V$

Step 2: Determine the current flowing through R2. The total current entering the parallel combination is $2.0 \, \text{A}$ (as shown by the current flowing through R1 and into the junction). By Kirchhoff's Current Law, the total current entering a junction equals the sum of currents leaving it. $I_{total parallel} = I_{4\Omega} + I_{R2}$ $2.0 \, A = 1.5 \, A + I_{R2}$ $I_{R2} = 2.0 \, A - 1.5 \, A$ $I_{R2} = 0.5 \, A$

Step 3: Apply Ohm's Law ($R = V/I$) to find the resistance of R2. $R_2 = \frac{V_{R2}}{I_{R2}}$ $R_2 = \frac{6.0 \, V}{0.5 \, A}$ $R_2 = 12 \, \Omega$ The resistance of resistor R2 is $\boxed{12 \Omega}$.

c) Calculate the resistance of resistor R1.

Step 1: Determine the potential difference across the 2.0 $\Omega$ resistor on the right. The current flowing through this resistor is $2.0 \, \text{A}$. $V_{2\Omega} = I_{2\Omega} \times R_{2\Omega}$ $V_{2\Omega} = 2.0 \, A \times 2.0 \, \Omega$ $V_{2\Omega} = 4.0 \, V$

Step 2: Determine the potential difference across R1 using Kirchhoff's Voltage Law. The total supply voltage is $12 \, \text{V}$. The components R1, the parallel combination, and the 2.0 $\Omega$ resistor on the right are in series. $V_{total} = V_{R1} + V_{parallel} + V_{2\Omega}$ We know $V_{total} = 12 \, \text{V}$, $V_{parallel} = 6.0 \, \text{V}$ (from part a), and $V_{2\Omega} = 4.0 \, \text{V}$. $12 \, V = V_{R1} + 6.0 \, V + 4.0 \, V$ $12 \, V = V_{R1} + 10.0 \, V$ $V_{R1} = 12 \, V - 10.0 \, V$ $V_{R1} = 2.0 \, V$

Step 3: Apply Ohm's Law ($R = V/I$) to find the resistance of R1. The current flowing through R1 is $2.0 \, \text{A}$. $R_1 = \frac{V_{R1}}{I_{R1}}$ $R_1 = \frac{2.0 \, V}{2.0 \, A}$ $R_1 = 1.0 \, \Omega$ The resistance of resistor R1 is $\boxed{1.0 \Omega}$.

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