Identify the series combination.

The $3 \, \Omega$ and $5 \, \Omega$ resistors are connected in series. Their equivalent resistance is:

The $8 \, \Omega$ series combination is in parallel with the $2 \, \Omega$ resistor, and this entire parallel arrangement is connected directly across the $6.0 \, V$ battery. Therefore, the voltage across the $8 \, \Omega$ series combination is $6.0 \, V$.

Since the $3 \, \Omega$ resistor is part of the $8 \, \Omega$ series combination, the current flowing through it is the total current through that branch. Using Ohm's Law ($I = \frac{V}{R}$):

The current through the $3 \, \Omega$ resistor is $\boxed{\text{0.75 A}}$.

2) If each cell has an internal resistance of $0.75 \, \Omega$, determine the total resistance in the circuit:

The $3 \, \Omega$ and $5 \, \Omega$ resistors are in series:

This $R_{s1}$ is in parallel with the $2 \, \Omega$ resistor. The equivalent resistance of this parallel combination is:

The battery has an EMF of $6.0 \, V$. Assuming standard $1.5 \, V$ cells, the number of cells in the battery is:

Each cell has an internal resistance of $0.75 \, \Omega$. So, the total internal resistance for 4 cells in series is:

The total resistance in the circuit is the sum of the total external resistance and the total internal resistance:

The total resistance in the circuit is $\boxed{\text{4.