The problem with your reasoning is that $\tan$ is invertible only when it is defined as a function $\tan: \left(-\frac{\pi}{2},\frac{\pi}{2} \right) \to \mathbb{R}$. Of course, it is invertible over other choice of domain too but this is the natural domain which is used. So, it follows that $\arctan: \mathbb{R} \to \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ is the inverse that you want. So, we see that:
$$\forall x \in \mathbb{R}: -\frac{\pi}{6} < \frac{\arctan(x)}{3} < \frac{\pi}{6}$$
Observe that $\cos$ is an even function. So, we can just consider the cases where $\frac{\arctan(x)}{3} \geq 0$. But now, notice that in the interval $\left[0,\frac{\pi}{6} \right)$, you have that $\cos$ is decreasing. Since $\cos$ is continuous, it follows that:
$$\text{Im}(f) = \left(\frac{\sqrt{3}}{2},1 \right]$$
As for your second argument, you have a decent idea but you need to elaborate further. Assume that $f$ is periodic with some period $p > 0$. Then, it follows that:
$$\forall x \in \mathbb{R}: f(x+p) = f(x)$$
But now, notice that:
$$f(0) = f(p) \implies 1 = \cos \left(\frac{\arctan(p)}{3} \right)$$
Using the information we have derived above about the range of $\arctan$, this implies that $\arctan(p) = 0$. But this means that $p = 0$ and that is a contradiction. It follows that $f$ is not periodic.
