Nyquist frequency mirroring
I understand the Nyquist frequency a bit differently - maybe this will help you.

There is only one Nyquist frequency, and it is at
$$
F_{n} = \frac{F_{s}}{2}
$$
Then, any signals with a frequency above \$+F_{n}\$ or below \$-F_{n}\$ are "folded" - they appear in the region \$[-F_{n}, F_{n}]\$. Here's an example:



Here's what's going on in this picture:


Zone 1 is inside the folding frequency, so the signal there isn't modified
Zone 2 is just above \$f_n\$, so it is flipped over to the other side. The aliased version of Zone 2 is identical to Zone 1.
Zone 3 is above \$f_n\$, so it is flipped over. When this happens, it appears on the negative side of the frequency axis - it is a mirror image of Zone 1.
Zone 4 follows the same process - it is flipped twice, putting it on top of Zone 1.


Now, I'll try to answer your questions:


\$F = 0\$ is not a folding frequency. See question 2.
The reason why you can see signals at, say, \$F = -3\text{ kHz}\$ is because we need both positive and negative frequencies to make signals. For example, a cosine at \$F = f_0\$ is
$$
\cos(2\pi f_0 t) = e^{i \cdot 2\pi f_0 t} + e^{-i \cdot 2\pi f_0 t}
$$
and the second term has the negative frequency you're seeing. It's impossible to make real (non-complex) signals without negative frequencies!
It sounds like you're describing this backwards. If you sample at \$F_s = 10 \text{ kHz}\$, then cosine signals at the frequencies
$$
F = 3, 7, 13, 17, 23, 27, \dots \text{ kHz}
$$
should all appear the same. Here's a quick plot that I made to show this:




This plot is \$\cos(3 \cdot 2\pi t)\$ and \$\cos(7 \cdot 2\pi t)\$. Notice that they intersect at 0, 0.1, 0.2, ..., so if you only sample at these times, you'll see the exact same signal.