Let the period be [tex]p\neq0[/tex]. Then by definition, a [tex]p[/tex]-periodic function [tex]f(x)[/tex] satisfies
[tex]f(x)=f(x+p)[/tex]
where here [tex]f(x)=\cos3x[/tex].
[tex]\cos3x=\cos3(x+p)=\cos(3x+3p)[/tex]
We know that [tex]\cos x[/tex] is [tex]2\pi[/tex]-periodic, i.e.
[tex]\cos x=\cos(x+2\pi)[/tex]
which reveals that we should also have
[tex]\cos3x=\cos(3x+2\pi)[/tex]
and so [tex]p[/tex] is such that [tex]3p=2\pi\implies p=\dfrac{2\pi}3[/tex].
In general, the period of [tex]\cos nx[/tex] is [tex]\dfrac{2\pi}n[/tex]. (The same applies for [tex]\sin nx[/tex].)
The range can be found by recalling that [tex]-1\le\cos3x\le1[/tex], which means [tex]0\le\cos3x+1\le2[/tex].
