The electromotive force E and the internal resistance r of a...

When the switch is thrown to the other side, the voltmeter is connected in parallel with the series combination of the resistor $R$ and the ammeter.

Resistance of the series combination of $R$ and the ammeter:
$R_s = R + R_A = 50 + 2.0 = 52 \text{ }\Omega$

Equivalent resistance of the parallel combination of the voltmeter and $R_s$:
$R_p = \dfrac{R_s R_V}{R_s + R_V} = \dfrac{52 \times 200}{52 + 200} = \dfrac{10400}{252} = \dfrac{2600}{63} \text{ }\Omega$

Total equivalent resistance of the circuit including the internal resistance of the battery:
$R_{eq} = R_p + r = \dfrac{2600}{63} + 1.0 = \dfrac{2663}{63} \text{ }\Omega$

Total current drawn from the battery:
$I = \dfrac{\mathcal{E}}{R_{eq}} = \dfrac{4.3}{\dfrac{2663}{63}} = \dfrac{270.9}{2663} \text{ A}$

The voltmeter reading is the potential difference across the parallel combination:
$V = I \times R_p = \dfrac{270.9}{2663} \times \dfrac{2600}{63} = \dfrac{4.3 \times 2600}{2663} = \dfrac{11180}{2663} \approx 4.2 \text{ V}$

The ammeter reading is the current flowing through the series combination of $R$ and the ammeter:
$I_A = \dfrac{V}{R_s} = \dfrac{\dfrac{11180}{2663}}{52} = \dfrac{215}{2663} \approx 0.08 \text{ A}$

Answer: $0.08 \text{ A}$, $4.2 \text{ V}$