Here's a crude way of estimating the answers:
The density of stars in the Milky Way disk decreases exponentially as you go out in radius and also exponentially as you go out in height above/below the disk midplane. The corresponding scale lengths are about 3 kpc (3 kiloparsecs = about 10,000 ly) radially and about 300 parsecs (about 1,000 ly) vertically ("scale height"); this means that if you go out in radius by about 10,000 ly, or up/down in height by about 1,000 ly, the density decreases by a factor of e (2.718...). Changing radius by about one tenth of the scale length or less, or changing height by about one tenth of the scale height or less, only changes the local density by about 10%, which isn't enough to worry about. That corresponds to about 300 parsecs (1000 ly) in radius, or about 100 ly in height.
So the for case of a 100-ly radius, you can assume the density everywhere is about the same as the local density suggested by the table you linked to. That density is 60 stars / ($(4/3) \, \pi \, 16.3^3$) = 0.0033 stars per cubic ly. For a sphere of radius 100 ly, you would then expect about 14,000 stars.
Naively, a sphere with a radius of 1000 ly would then have about 14 million stars. But a sphere that big extends above and below the disk midplane enough so that it samples about 1 scale height. So you can't just re-use the local density. Without getting into the proper approach of integrating the density functions in 3D, we can make a crude estimate as follows: If we used the density at one scale height (1/e times the local density), then the sphere would have only about 4.5 million stars. If we split the difference, that's about 10 million stars, which is probably not too terribly wrong. (I've been ignoring things like stars belonging to the "thick disk" and stars belonging to the halo, but they won't contribute a whole lot; also, the Sun isn't exactly at the disk midplane, but about 60 or 70 ly above it...)