Figure 16: Circuit Analysis

$$
\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}
$$
$$
\frac{1}{R_p} = \frac{1}{2 \, \Omega} + \frac{1}{3 \, \Omega} = \frac{3 + 2}{6 \, \Omega} = \frac{5}{6 \, \Omega}
$$
$$
R_p = \frac{6}{5} \, \Omega = 1.2 \, \Omega
$$

The parallel combination ($R_p$) is in series with the $5 \, \Omega$ resistor ($R_3$).
$$
R_{ext} = R_p + R_3
$$
$$
R_{ext} = 1.2 \, \Omega + 5 \, \Omega = 6.2 \, \Omega
$$

$$
I_{total} = \frac{V}{R_{ext}}
$$
$$
I_{total} = \frac{6.0 \, \text{V}}{6.2 \, \Omega} = \frac{60}{62} \, \text{A} = \frac{30}{31} \, \text{A}
$$

$$
V_p = I_{total} \times R_p
$$
$$
V_p = \frac{30}{31} \, \text{A} \times 1.2 \, \Omega = \frac{30}{31} \times \frac{6}{5} \, \text{V} = \frac{6 \times 6}{31} \, \text{V} = \frac{36}{31} \, \text{V}
$$

$$
I_{3\Omega} = \frac{V_p}{3 \, \Omega}
$$
$$
I_{3\Omega} = \frac{\frac{36}{31} \, \text{V}}{3 \, \Omega} = \frac{36}{31 \times 3} \, \text{A} = \frac{12}{31} \, \text{A}
$$
$$
I_{3\Omega} \approx 0.387 \, \text{A}
$$
The current through the $3 \, \Omega$ resistor is $\boxed{\text{0.387 A}}$.

From part (d), $R_{ext} = 6.2 \, \Omega$.

$$
R_{total} = R_{ext} + r
$$
$$
R_{total} = 6.2 \, \Omega + 0.75 \, \Omega
$$
$$
R_{total} = 6.95 \, \Omega
$$
The total resistance in the circuit is $\boxed{\text{6.95 \Omega}}$.