Calculate the current delivered by the battery, the emf (E) of the battery, and write down the resistance of N with a reason.

The effective resistance between points a and b is $R_{ab} = 6 \, \Omega$.
The voltmeter $V_1$ reads the potential difference across points a and b, so $V_{ab} = 5 \, V$.

This current is the main current delivered by the battery.

The current delivered by the battery is $\boxed{\text{0.83 A}}$.

9.2.2 Emf ($\mathcal{E}$) of the battery

The external circuit consists of the effective resistance between a and b ($R_{ab}$) in series with resistor T ($R_T$).

The EMF of the battery is $\boxed{\text{7 V}}$.

9.2.3 Resistance of N

The resistance of N is $\boxed{\text{3 \Omega}}$.
Reason: Resistors M and N are connected in series between points a and b. The total voltage across this series combination is $V_{ab} = 5 \, V$. Voltmeter $V_2$ measures the voltage across resistor M, which is $V_M = 2.5 \, V$. In a series circuit, the sum of individual voltages equals the total voltage, so $V_N = V_{ab} - V_M = 5 \, V - 2.5 \, V = 2.5 \, V$. Since the voltage across M ($V_M = 2.5 \, V$) is equal to the voltage across N ($V_N = 2.5 \, V$), and the current through series resistors is the same, their resistances must be equal ($R_M = R_N$). Given that the effective resistance of the series combination is $R_{ab} = R_M + R_N = 6 \, \Omega$, it follows that $R_N = \frac{R_{ab}}{2} = \frac{6 \, \Omega}{2} = 3 \, \Omega$.