When the switch is closed, calculate: (i) the combined resistance of the circuit; (ii) the change in the current in the circuit when the 4.00 Ω resistor is disconnected.

The resistors $R_2 = 2.00 \, \Omega$, $R_3 = 4.00 \, \Omega$, and $R_4 = 2.00 \, \Omega$ are in parallel.

$$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$$ $$\frac{1}{R_p} = \frac{1}{2.00 \, \Omega} + \frac{1}{4.00 \, \Omega} + \frac{1}{2.00 \, \Omega}$$ $$\frac{1}{R_p} = 0.500 \, \Omega^{-1} + 0.250 \, \Omega^{-1} + 0.500 \, \Omega^{-1}$$ $$\frac{1}{R_p} = 1.250 \, \Omega^{-1}$$ $$R_p = \frac{1}{1.250 \, \Omega^{-1}} = 0.800 \, \Omega$$

The 6.00 $\Omega$ resistor is in series with the parallel combination $R_p$.

$$R_{total,1} = 6.00 \, \Omega + R_p$$ $$R_{total,1} = 6.00 \, \Omega + 0.800 \, \Omega$$ $$R_{total,1} = \boxed{6.80 \, \Omega}$$

The voltage supply is $V = 12.0 \, \text{V}$.

$$I_1 = \frac{V}{R_{total,1}}$$ $$I_1 = \frac{12.0 \, \text{V}}{6.80 \, \Omega} \approx 1.7647 \, \text{A}$$

Now, only $R_2 = 2.00 \, \Omega$ and $R_4 = 2.00 \, \Omega$ are in parallel.

$$\frac{1}{R_p'} = \frac{1}{R_2} + \frac{1}{R_4}$$ $$\frac{1}{R_p'} = \frac{1}{2.00 \, \Omega} + \frac{1}{2.00 \, \Omega}$$ $$\frac{1}{R_p'} = 0.500 \, \Omega^{-1} + 0.500 \, \Omega^{-1}$$ $$\frac{1}{R_p'} = 1.000 \, \Omega^{-1}$$ $$R_p' = \frac{1}{1.000 \, \Omega^{-1}} = 1.00 \, \Omega$$

$$R_{total,2} = 6.00 \, \Omega + R_p'$$ $$R_{total,2} = 6.00 \, \Omega + 1.00 \, \Omega = 7.00 \, \Omega$$ $$I_2 = \frac{V}{R_{total,2}}$$ $$I_2 = \frac{12.0 \, \text{V}}{7.00 \, \Omega} \approx 1.7143 \, \text{A}$$

$$\Delta I = I_2 - I_1$$ $$\Delta I = 1.7143 \, \text{A} - 1.7647 \, \text{A}$$ $$\Delta I = \boxed{-0.0504 \, \text{A}}$$

The current decreases by 0.0504 A.

The motorcycle battery (12.0 V) is charging the phone battery (4.6 V). The phone battery's EMF opposes the charging voltage.

$$V_{net} = V_{motorcycle} - V_{phone}$$ $$V_{net} = 12.0 \, \text{V} - 4.6 \, \text{V}$$ $$V_{net} = 7.4 \, \text{V}$$

Using Ohm's Law, $V_{net} = I \times R_{total\_bulbs}$, where $I = 0.8 \, \text{A}$.

$$R_{total\_bulbs} = \frac{V_{net}}{I}$$ $$R_{total\_bulbs} = \frac{7.4 \, \text{V}}{0.8 \, \text{A}}$$ $$R_{total\_bulbs} = 9.25 \, \Omega$$

Since the two bulbs are in series, their resistances add up.

$$R_{each\_bulb} = \frac{R_{total\_bulbs}}{2}$$ $$R_{each\_bulb} = \frac{9.25 \, \Omega}{2}$$ $$R_{each\_bulb} = \boxed{4.625 \, \Omega}$$