Determining output of a LTI system
You are right that the system $h[n] = 2^n.u[n]$ is unstable in absolute summability sense. And absolute summability of impulse response of an LTI system is the required condition for BIBO stability. Meaning Bounded Input bounded output stability. Here the input is bounded and it is only corresponding to 4 frequencies in digital frequency domain $\omega = -\pi /3, \pi /3, -\pi/2, \pi/2$. The magnitude of input cannot be greater than 3 at any value of $n$. But, the output will not be bounded because the system is not BIBO stable.

Even if you start the system at 0, meaning $y[n] = 0, \forall n &lt; 0$, $y[n]$ will blow up as $n$ grows.
$$y[n] = \sum^{\infty}_{m=0}2^m.x[n-m], \forall n \ge 0$$
Putting $x[n] = cos[\frac{\pi n}{3}]$ as input:
$$y[n] = \sum^{\infty}_{m=0}2^m. \frac{e^{j\pi (n-m)/3} + e^{-j\pi (n-m)/3}}{2}$$
$$y[n] = \frac{e^{j\pi n/3}}{2}.\sum^{\infty}_{m=0}(2.e^{-j\pi/3})^m + \frac{e^{-j\pi n/3}}{2}.\sum^{\infty}_{m=0}(2.e^{j\pi/3})^m$$
Since both the sums are going to blow up as $n$ grows, we can say that $y[n]$ is not bounded.

Another way of saying this is that $H(e^{j\omega})$ does not exist for this system ,because DTFT sum does not converge. And so we cannot get $|H(e^{j\omega})|$ and $\angle{H(e^{j\omega})}$ for any $\omega \in [-\pi, \pi]$.

If we considered the system to be anti-causal, then the unit circle will fall inside ROC because ROC will become $|z|&lt;2$. And the anti-causal but stable system's impulse response will be given by : $h[n] = 2^n.u[-n]$. 

But this will be wrong thing to do, because the difference equation represents a Causal System, since the current output sample depends only on current input sample and a past output sample.