Non-unique Sample Rate
You're doing something wrong with the additional \$2\pi\$. You might just be forgetting that \$cos(x + 2\pi) = cos(x)\$. I assume you are leaving it in and ignoring the order of operations (bad parenthesis).

$$\cos\left(\frac{\pi}{3} n + 2\pi\right) = \cos\left(\frac{\pi}{3} n\right)\require{cancel}$$

So, using your steps and adding the \$2\pi\$.

$$\cos\left(\frac{\pi}{3} n \cancel{+ 2\pi}\right) = \cos(4000\pi nT\cancel{+ 2\pi})$$
$$\cos\left(\frac{\pi}{3} n \right) = \cos(4000\pi nT)$$
$$\frac{\pi}{3} n = 4000\pi nT$$
$$T = \frac{1}{12000}$$

It's a pretty simple mistake, thus rather easy to make.



Edit:


  if we shift ONLY x[n] and not x_c(t), then T changes


$$\cos\left(\frac{\pi}{3} n + 2\pi\right) = \cos(4000\pi nT)$$
$$\cos\left(\frac{\pi}{3} n \cancel{+ 2\pi}\right) = \cos(4000\pi nT)$$
$$\cos\left(\frac{\pi}{3} n \right) = \cos(4000\pi nT)$$
$$\frac{\pi}{3} n = 4000\pi nT$$
$$T = \frac{1}{12000}$$

No, it doesn't.



Edit 2: 


  $$\cos\left(\frac{\pi}{3} n + 2\pi\right) = \cos(4000\pi nT)$$
  $$\frac{\pi}{3} n + 2\pi = 4000\pi nT$$


Ok, the problem is you're ignoring the sum angle in your cosine when canceling it. Trigonometry doesn't work that way. You can leave the \$2\pi\$ in, apply the proper identity, and cancel the result. Or, you can simply cross off any multiple of \$2\pi\$ and proceed as you were already doing.