A) When a negatively charged electroscope is touched with a finger, the excess electrons on the electroscope are repelled by each other and flow through the human body to the Earth. This process,

(ii) The total charge ($Q_{total}$):

The total charge is given by $Q_{total} = C_{total} \times V$.

The total charge is $\boxed{\text{36 µC}}$.

(iii) Voltage across 4µF ($V_4$):

The voltage across the parallel combination ($C_{5-14}$) is equal to the battery voltage because it is in parallel with the 2 µF capacitor, and this entire parallel arrangement is connected directly across the battery.
So, the voltage across the series branch containing $C_5$ and $C_{14}$ is $V = 8.0 \, \text{V}$.
The charge in this series branch is $Q_{5-14} = C_{5-14} \times V$.

This charge is the same for both $C_5$ and the parallel combination $C_{14}$. So, $Q_{14} = 20 \, \mu\text{C}$.
Now, find the voltage across $C_{14}$:

Since the 1 µF and 4 µF capacitors are in parallel, the voltage across them is the same as $V_{14}$.
Therefore, the voltage across the 4 µF capacitor is $V_4 = V_{14}$.

The voltage across the 4µF capacitor is $\boxed{\text{4.0 V}}$.

(iv) Energy stored in 4µF ($E_4$):

The energy stored in a capacitor is given by $E = \frac{1}{2} C V^2$.
Using $C_4 = 4 \, \mu\text{F}$ and $V_4 = 4.0 \, \text{V}$:

The energy stored in the 4µF capacitor is $\boxed{\text{32 µJ}}$.