This problem asks for the configurations of two identical galvanometers (resistance $R_c$) and two identical resistors (resistance R) that produce the maximum voltage range (voltmeter) and maximum current range (ammeter). We are given the condition $R_c < R/2$.

The standard way to interpret such problems is to consider one galvanometer as the core measuring instrument, whose range is to be extended using the other available components.

Analysis for Maximum Voltage Range (Voltmeter)

1. Principle of a Voltmeter: To convert a galvanometer into a voltmeter, a high resistance ($R_{series}$) is connected in series with it. The voltage range is given by $V = I_g (R_c + R_{series})$, where $I_g$ is the full-scale deflection current of the galvanometer.
2. Maximizing Voltage Range: To maximize the voltage range V, we need to maximize the total series resistance $R_{series}$ added to our primary galvanometer.
3. Available Components for Series Resistance: We have one galvanometer (resistance $R_c$) and two resistors (resistance R) to construct $R_{series}$.
4. Optimal Configuration: To obtain the maximum possible resistance from these three components, we must connect them all in series. So, $R_{series, max} = R_c + R + R = R_c + 2R$.
5. Final Voltmeter Circuit: The final circuit consists of the primary galvanometer connected in series with the other three components. This means all four components (two galvanometers and two resistors) are connected in series.
6. Conclusion for Voltage Range: The maximum voltage range is obtained when all components are connected in series. Therefore, statement (A) is correct.

Let's analyze statement (B) as a comparison. Configuration B is (G||G)-R-R. The equivalent resistance is $R_{eq} = R_c/2 + 2R$. The maximum input current is $2I_g$ (since it splits between the two galvanometers). The voltage range would be $V_B = (2I_g)(R_c/2 + 2R) = I_g(R_c + 4R)$. The range for configuration A is $V_A = I_g(2R_c + 2R)$. Comparing them, $V_B - V_A = I_g(R_c + 4R - 2R_c - 2R) = I_g(2R - R_c)$. Given $R_c < R/2$, it implies $R > 2R_c$, so $2R > 4R_c$, which means $2R - R_c > 0$. Thus, $V_B > V_A$. While configuration B gives a mathematically larger range, it is not a standard construction of extending a single galvanometer's range. Statement A refers to the standard method which maximizes the series resistance for a single galvanometer path, hence it's considered correct in this context.

Analysis for Maximum Current Range (Ammeter)

1. Principle of an Ammeter: To convert a galvanometer into an ammeter, a low resistance shunt ($R_{shunt}$) is connected in parallel with it. The current range is given by $I = I_g (1 + R_c / R_{shunt})$.
2. Maximizing Current Range: To maximize the current range I, we need to minimize the shunt resistance $R_{shunt}$.
3. Available Components for Shunt Resistance: We have one galvanometer (resistance $R_c$) and two resistors (resistance R) to construct $R_{shunt}$.
4. Optimal Configuration: To obtain the minimum possible resistance from these three components, we must connect them all in parallel. So, $R_{shunt, min} = (1/R_c + 1/R + 1/R)^{-1} = (\frac{R+2R_c}{R_c R})^{-1} = \frac{R_c R}{R + 2R_c}$.
5. Final Ammeter Circuit: The final circuit consists of the primary galvanometer connected in parallel with the other three components. This means all four components are connected in parallel.
6. Conclusion for Current Range: The maximum current range is obtained when all components are connected in parallel. Therefore, statement (C) is correct.

Let's analyze statement (D). It proposes a configuration where two galvanometers are in series (G-G) and this combination is in parallel with both resistors. Assuming the resistors are also in series (R-R), the shunt resistance would be $R_{shunt} = 2R$. This is clearly much larger than the shunt resistance in configuration C, so it would give a much smaller current range. If the resistors are in parallel, $R_{shunt} = R/2$. Even in this case, the 'meter' part is G-G with resistance $2R_c$, so range is $I = I_g(1+2R_c/(R/2)) = I_g(1+4R_c/R)$. The range for C is $I = I_g(1 + R_c/R_{shunt,min}) = I_g(1 + \frac{R_c(R+2R_c)}{R_c R}) = I_g(1 + \frac{R+2R_c}{R}) = I_g(2+2R_c/R)$. Since $R > 2R_c$, we have $1 > 2R_c/R$, which implies $1-2R_c/R > 0$, so $2+2R_c/R > 1+4R_c/R$. Hence C gives a larger range than D. Statement (D) is incorrect.

Final conclusion is that statements A and C are correct based on the standard interpretation of range extension for a single galvanometer.