$x[n]$ after sampling of $cos(16\pi t+\phi)$ at 12kHz I'm not sure what the question really means, so this is just guesswork.

I think options 1 and 4 can be ruled out as $w_0&lt;\pi$.

The CTFT of $cos(16\pi t+\phi)$ has two spikes at $16\pi$ and $-16\pi$. I guess that the question means that we sampled this spectrum at $12Hz$, (because 12kHz doesn't make sense to me, might be a typo in the question), i.e. at $24\pi$.

So we get one copy of $-16\pi$ at $-16\pi+24\pi=8\pi$. Similarly we get a copy of $16\pi$ at $-8\pi$.

Now we normalize to the $w$ by diving by $12$. Only the frequencies $-8\pi$ and $8\pi$ get mapped in the range $-\pi$ yo $\pi$ as required by the question. These frequencies get mapped at $-\frac{2}{3}\pi$ and $\frac{2}{3}\pi$. But the problem is that they also get scaled by $12Hz$ during sampling (and I'm ignoring that to get to one of the options). So the $x[n]$ corresponding to $\frac{2}{3}\pi$ and $frac{2}{3}\pi$ is $cos(\frac{2}{3}\pi n+\phi)$

Is this correct?
First of all, as you said the sampling rate is probably $12$ Hz, rather than $12$ kHz, and perhaps they want to demonstrate an aliasing example.

Given a bandlimited continuous-time periodic signal 
$$x(t) = \cos(16\pi t + \phi)$$ 
the samples taken at the rate $F_s = 12$ Hz will be denoted as $x[n]$ and will be obtained by via $x[n] = x(t_n)$ with $t_n = n T_s = n/Fs$ :

$$x[n] = \cos( 16 \pi \frac{1}{12} n + \phi) = \cos(\frac{4 \pi}{3} n + \phi)$$

$x[n]$ is a discrete-time periodic sequence with frequency $w_0 = \frac{4\pi}{3}$, however, in the discrete-time case we are interested in the range of frequencies that fall in $ [-\pi,\pi)$, and therefore this signal's frequency will be cast as:

$$ x[n] = \cos(\frac{4 \pi}{3} n + \phi) = \cos(\frac{2 \pi}{3} n + \phi)$$

which indicates that the original continuous-time frequency of $8$ Hz is lost, and a new aliased frequency of $4$ Hz is attained...