Electronic – ny simple way or trick to express given signal using waveforms of sum of simple sinusoidal signals

Question 1: Assuming that the oscillator (perhaps programmable) is given, what kind of hardware would you suggest for (2) and (3) ?

For 2: All-pass (phase-shifting) filter, a.k.a. Hilbert transformer

For 3: Balanced modulator

Question 2: Does this setup have significant shortcomings (beside cost of components?)

Yes. While direct conversion to complex baseband is regularly done in the digital domain, it can be quite finicky to get the same concept working well (and reliably) in the analog domain. It's nearly always easier to do the detection at an intermediate RF frequency.

Question 3: is this observation [about square-wave local oscillators] correct?

Yes, except that you want a bandbass filter that's centered on the carrier frequency, not a lowpass filter. Sometimes this filter has a bandwidth of approximately 2B, in which case it needs to be tuned along with the local oscillator. This sort of setup is called a preselector, and is commonly used in superheterodyne receivers, such as those used for the AM and FM broadcast bands.

But sometimes, a fixed bandpass filter that covers the entire band of interest is used instead. This works as long as none of the unwanted "image" bands ever falls into the band of interest. This is more common in 2-way VHF communications systems, such as those used for air traffic control and public service (police, fire).

For any periodic signal you get

$$\int_{-\infty}^{\infty}|x(t)|^2dt=\infty$$

i.e. the integral does not converge, and, consequently, the energy is infinite. The power is finite and can be computed from the following formula (which differs from yours by a factor of \$\frac12\$):

$$\overline{x^2(t)}=\frac{1}{2T}\int_{-T}^T|x(t)|^2dt$$

Note that \$\cos^2 x=\frac12 (1+\cos(2x))\$ and \$\sin^2 x=\frac12 (1-\cos(2x))\$. So after integrating over one period, the terms with double frequency and also the cross-term cancel out. So you finally get for the power

$$\overline{x^2(t)}=10^2\cdot \frac12 + 5^2\cdot\frac12=62.5$$

Note that the actual value of \$T\$ is irrelevant. This is a good thing because the value you got is wrong.