First, find the equivalent resistance of the parallel part: $$R_p = (1/3 + 1/6)^{-1} = (3/6)^{-1} = 2,\Omega$$. The total equivalent resistance is $$R_{eq} = 2,\Omega + R_p = 2,\Omega + 2,\Omega = 4,\Omega$$. The total current from the battery is $$I_{total} = V/R_{eq} = 12,\text{V} / 4,\Omega = 3,\text{A}$$. The voltage across the parallel part is $$V_p = I_{total} \times R_p = 3,\text{A} \times 2,\Omega = 6,\text{V}$$. The current through the $$3,\Omega$$ resistor is $$I_3 = V_p / 3,\Omega = 6,\text{V} / 3,\Omega = 2.0,\text{A}$$.

When the four resistors are in series, the equivalent resistance is $$R_S = R+R+R+R = 4R$$. The current is $$I_S = \mathcal{E}/(4R)$$. When the four resistors are in parallel, the equivalent resistance is $$R_P = (1/R + 1/R + 1/R + 1/R)^{-1} = (4/R)^{-1} = R/4$$. The current is $$I_P = \mathcal{E}/(R/4) = 4\mathcal{E}/R$$. The ratio is $$I_P / I_S = (4\mathcal{E}/R) / (\mathcal{E}/(4R)) = 16$$.

Let the total current be $$I$$. Power in $$R_1$$ is $$P_1 = I^2 R_1 = 4I^2$$. The equivalent resistance of the parallel part is $$R_p = (1/12 + 1/6)^{-1} = 4,\Omega$$. The voltage across the parallel part is $$V_p = I R_p = 4I$$. The power in $$R_2$$ is $$P_2 = V_p^2 / R_2 = (4I)^2 / 12 = 16I^2/12 = (4/3)I^2$$. The power in $$R_3$$ is $$P_3 = V_p^2 / R_3 = (4I)^2 / 6 = 16I^2/6 = (8/3)I^2$$. Comparing the powers: $$P_1 = 4I^2$$, $$P_2 \approx 1.33I^2$$, $$P_3 \approx 2.67I^2$$. Therefore, the correct ranking is $$P_1 > P_3 > P_2$$.

For the nonideal battery, $$I = \mathcal{E} / (R_{ext} + r)$$. For the ideal battery, $$I_{ideal} = \mathcal{E} / R_{ext}$$. From the first equation, we can write $$E = I(R_{ext} + r)$$. Substituting this into the second equation gives $$I_{ideal} = I(R_{ext} + r) / R_{ext}$$. Distributing the denominator gives $$I_{ideal} = I(R_{ext}/R_{ext} + r/R_{ext}) = I(1 + r/R_{ext})$$.

The current $$I_1$$ flows through $$R_1$$ and is the total current for the circuit. This current then splits to flow through the parallel branches containing $$R_2$$ and $$R_3$$. By Kirchhoff's junction rule, $$I_1 = I_2 + I_3$$, so $$I_1$$ is the largest current. In a parallel combination, the potential difference across each branch is the same. Since $$I = V/R$$, the branch with the lower resistance will have the higher current. Because $$R_2 < R_3$$, it follows that $$I_2 > I_3$$. Therefore, the correct ranking is $$I_1 > I_2 > I_3$$.

The voltmeter reading gives the potential difference across the $$5,\Omega$$ resistor. The current through this resistor (which is the total current) is $$I = V/R = 15,\text{V} / 5,\Omega = 3,\text{A}$$. Next, find the equivalent resistance of the parallel branch: $$R_p = (1/10 + 1/30)^{-1} = (3/30 + 1/30)^{-1} = (4/30)^{-1} = 7.5,\Omega$$. The potential difference across the parallel branch is $$V_p = I R_p = 3,\text{A} \times 7.5,\Omega = 22.5,\text{V}$$. The EMF of the battery is the sum of the potential differences across the series components: $$EMF = V_5 + V_p = 15,\text{V} + 22.5,\text{V} = 37.5,\text{V}$$.

A real ammeter has a small but nonzero internal resistance. When placed in series in the circuit, this resistance adds to the existing resistance $$R$$. This increases the total equivalent resistance of the circuit. According to Ohm's law, $$I = V/R_{total}$$, an increase in the total resistance will cause a decrease in the total current. Therefore, the measured current will be less than the current measured by the ideal ammeter.

A real voltmeter is connected in parallel with resistor $$R_1$$. This creates a parallel combination with an equivalent resistance $$R_{p1}$$ that is less than $$R_1$$. This decrease in resistance for one part of the series circuit leads to a decrease in the total equivalent resistance of the entire circuit. According to the voltage divider principle, the fraction of the total voltage that drops across the $$R_1$$-voltmeter combination will be smaller than the fraction that dropped across $$R_1$$ alone. Thus, $$V_{real} < V_{ideal}$$.

The equivalent resistance of the parallel combination of $$R_2$$ and $$R_3$$ is $$R_p = (\frac{1}{R_2} + \frac{1}{R_3})^{-1} = \frac{R_2 R_3}{R_2 + R_3}$$. This combination is in series with $$R_1$$, so the total equivalent resistance is found by adding their resistances: $$R_{eq} = R_1 + R_p = R_1 + \frac{R_2 R_3}{R_2 + R_3}$$.

The equivalent resistance of two $$4,\Omega$$ resistors in parallel is $$R_p = (1/4 + 1/4)^{-1} = 2,\Omega$$. The total circuit resistance is the sum of the series resistor and this parallel equivalent: $$R_{eq} = 2,\Omega + 2,\Omega = 4,\Omega$$. The total current is $$I = V/R_{eq} = 12,\text{V} / 4,\Omega = 3,\text{A}$$. The potential difference across the parallel combination is $$V_p = I \times R_p = 3,\text{A} \times 2,\Omega = 6,\text{V}$$.