If someone asked for the average power dissipated in a device, what
would that mean?
The average power is the time average of the instantaneous power. In the case you describe, the instantaneous power is a 1W peak square wave and, as you point out, the average over a period is zero.
But, consider the case of (in phase) sinusoidal voltage and current:
$$v(t) = V \cos \omega t $$
$$i(t) = I \cos \omega t $$
The instantaneous and average power are:
$$p(t) = v(t) \cdot i(t) = V_m \cos\omega t \cdot I_m \cos\omega t = \dfrac{V_m \cdot I_m}{2}(1 + \cos2\omega t) $$
$$p_{avg} = \dfrac{V_m \cdot I_m}{2}$$
(since the time average of sinusoid over a period is zero.)
In the above, we evaluated the time average of the instantaneous power. This will always give the correct result.
You link to the Wiki article on AC power which is analyzed in the phasor domain. Phasor analysis assumes sinusoidal excitation so it would be a mistake to apply the AC power results to your square wave example.
The product of the rms phasor voltage \$\vec V \$ and current \$\vec I \$ gives the complex power S:
$$S = \vec V \cdot \vec I = P + jQ$$
where P, the real part of S, is the average power.
The rms phasor voltage and current for the time domain voltage and current above are:
$$\vec V = \dfrac{V_m}{\sqrt{2}} $$
$$\vec I = \dfrac{I_m}{\sqrt{2}} $$
The complex power is then:
$$S = \dfrac{V_m}{\sqrt{2}}\dfrac{I_m}{\sqrt{2}} = \dfrac{V_m \cdot I_m}{2}$$
Since, in this case, S is purely real, the average power is:
$$P = \dfrac{V_m \cdot I_m}{2}$$
which agrees with the time domain calculation.
The source of the confusion here is from the second equation you list:
$$P_\mathrm{avg2} = V_\mathrm{rms}\times I_\mathrm{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} V^2(t) \, \mathrm{d}t} \times \sqrt{\frac{1}{T} \int_{0}^{T} I^2(t) \, \mathrm{d}t}$$
This equation is true under the assumption that both the voltage and currents are sinusoidal and in phase. The derivation works out that way, so it is a simple way to calculate average power under that specific circumstance. The instantaneous method you used is the general case, and should work under all circumstances.
Best Answer
You have made a false math assumption. You have caught from the wind "taking average and multiplication are distributive". That's not true as you have already found. You can check it with two voltage samples U1, U2 and two current samples I1, I2
The average power is (U1*I1 + U2*I2)/2. There's no way to reduce this to ((U1+U2)/2)*((I1+I2)/2)
You must calculate P(t)=U(t)*I(t). That's the momentary power. The average power is the average of U(t)*I(t) calculated over the period of interest. With sinusoidal current and voltage we calculate the average over one cycle.