Calculate the combined resistance of the 2 and 3 resistors.

These two resistors are connected in parallel.

$$R_{p} = \frac{R_1 \times R_2}{R_1 + R_2}$$ $$R_{p} = \frac{2\,\Omega \times 3\,\Omega}{2\,\Omega + 3\,\Omega}$$ $$R_{p} = \frac{6\,\Omega^2}{5\,\Omega}$$ $$R_{p} = 1.2\,\Omega$$

The combined resistance of the $2\,\Omega$ and $3\,\Omega$ resistors is $\boxed{\text{1.2\,\Omega}}$.

The equivalent resistance of the parallel combination ($R_p = 1.2\,\Omega$) is in series with the $6\,\Omega$ resistor.

$$R_{total} = R_p + R_{series}$$ $$R_{total} = 1.2\,\Omega + 6\,\Omega$$ $$R_{total} = 7.2\,\Omega$$

The total resistance in the circuit when the switch is closed is $\boxed{\text{7.2\,\Omega}}$.

First, calculate the total current ($I_{total}$) flowing from the $12\,V$ source using Ohm's Law.

$$I_{total} = \frac{V_{source}}{R_{total}}$$ $$I_{total} = \frac{12\,V}{7.2\,\Omega}$$ $$I_{total} = \frac{120}{72}\,A$$ $$I_{total} = \frac{5}{3}\,A$$

The potential difference across the parallel combination ($V_p$) is the product of the total current and the equivalent parallel resistance. The $2\,\Omega$ resistor is part of this parallel combination, so the potential difference across it is $V_p$.

$$V_p = I_{total} \times R_p$$ $$V_p = \frac{5}{3}\,A \times 1.2\,\Omega$$ $$V_p = \frac{5}{3}\,A \times \frac{6}{5}\,\Omega$$ $$V_p = 2\,V$$

The potential difference across the $2\,\Omega$ resistor is $\boxed{\text{2\,V}}$.