How big should a lens of a telescope be so we can see the American flag on the Moon from the Earth's surface? How big in diameter should the lens of the imaginary telescope to be so we can see the American flag from e.g. some observatory in Hawaii?
 A: We are fortunate that NASA released some photos from the Lunar Reconnaissance Observatory (LRO) last year! Even from a mere 31 miles (50 km) up, it is impossible to see the actual flag, though we can see its shadow.

A: Assuming the resolution is limited only by diffraction, e.g. not by atmospheric turbulence, then the angular resolution is given by:
$$ \sin \theta = 1.22 \frac{\lambda}{D} $$
where $D$ is the telescope aperture. If the distance is $d_m$ and the size of the flag (or the feature on the flag you want to resolve) is $d_f$, then $\sin \theta = d_f/d_m$ so:
$$ d_f/d_m = 1.22 \frac{\lambda}{D} $$
or 
$$ D = 1.22 \lambda \frac{d_m}{d_f} $$
If you put the size of the flag as 1 metre the corresponding aperture size is about 240m.
A: You dont need a single large mirror to resolve small objects. You can use two mirrors separated by a distance as an interferometer. With adaptive optics it is possible to overcome atmospheric distortion.
The Keck pair of telescopes use this technique to get an effective mirror size of 85m, equal to the distance between the two mirrors. The resolution achieved is 5 milliarcseconds (mas) at 2.2 micron wavelength. (The VLT array of four telescopes resolves 3.5 mas) 
The flag on the moon subtends an angle of 0.5 mas, so if the resolution and adaptive optics could be scaled up it should be possible to see the flag with two mirrors separated by something like 850m. This is more than the other answers but I think it is more realistic taking into account the adaptive optics. 
A: The angular resolution of a telescope is approximated by the formula:
$$\sin \theta \approx\theta\approx1.220\frac{\lambda}{D}$$
So, if we know the angle, we can calculate the diameter $D$.
The rest really depends on how big the flag is and where on moon it has been planted. Assuming 50$\text {cm}$ as the diameter of the flag, and assuming the flag is planted somewhere perpendicular to earth's view line, and also doing it when moon is in its perigee($d=363,295\text{km}$); the diameter would be(for visible blue light):
$$D\approx\frac{1.22 \lambda}{\theta}\approx\frac{1.22 \times450\text{nm}}{\frac{50\text{cm}}{363295\text{km}}}\approx400\text{m}$$
If you don't trust my numbers, you may look at this link.
