To determine the output power of a class A large signal amplifier, we must delve into the concepts of how these amplifiers work with sinusoidal signals and load resistance. The class A amplifier operates linearly over the entire cycle of the input signal, meaning it amplifies both halves of a sine wave.

Let's analyze the formulas presented in the options and determine which correctly represents the output power.

1. \(\dfrac{(V_{max}-V_{min})(I_{max}-I_{min})}{2\sqrt{2}}\): This expression does not accurately reflect the correct formula for power in terms of voltage and current range differences.
2. \(\dfrac{(V_{max}-V_{min})(I_{max}-I_{min})}{8}\): This is the correct expression for the output power in a class A large signal amplifier. The calculation involves halving the peak power, which is appropriate due to the nature of the AC waveform symmetry around zero.
3. \(\dfrac{V_m\cdot I_m}{\sqrt{2}}\): This expression represents apparent power or RMS calculations, but it's not specific to the derived power from the max-min range of a class A amplifier.
4. \(\dfrac{(V_{max}-V_{min})(I_{max}-I_{min})}{2}\): While seemingly intuitive, this expression does not account for the AC waveform's averaging effect over a full cycle.

In class A amplifiers, power is calculated considering the linear operation over the full signal cycle. The significant loss component in such amplifiers due to both maximum and minimum voltages and currents leads us to compute the average over the waveform span:

Therefore, the correct expression for the output power of the amplifier given the peak sinusoidal voltage and current differences is:

\(\dfrac{(V_{max}-V_{min})(I_{max}-I_{min})}{8}\).

This option accounts for the waveform's entire cycle and provides the accurate average power assumed in the design and functioning of a class A amplifier.