No. If $x = 3^n \pi$ then $|f(x)| = f(\pi) \neq 0$ (numerically it's about $0.466$), and $3^n \pi$ can be arbitrarily large.

Such a construction fails for $\prod_{k=1}^\infty \cos(x/2^k)$ because if some $x/2^k$ is nearly $\pi$, or more generally some odd multiple of $\pi$, then $x/2^{k+1}$ is nearly a half-integral multiple of $\pi$ and thus has a very small cosine.

No. If $x = 3^n \pi$ then $|f(x)| = f(\pi) \neq 0$ (numerically it's about $0.466$), and $3^n \pi$ can be arbitrarily large.

Such a construction fails for $\prod_{k=1}^\infty \cos(x/2^k)$ because if some $x/2^k$ is nearly $\pi$, or more generally some odd multiple of $\pi$, then $x/2^{k+1}$ is nearly a half-integral multiple of $\pi$ and thus has a very small cosine.

[added later] For $f_a(x) := \prod_{k=1}^\infty \cos(x/a^k)$ with $a > 2$, the same construction does work if $a$ is an integer (consider $x = a^n \pi$), and more generally if $a$ is a Pisot-Vijayaraghavan number (which need not exceed $2$, e.g. if $a = (1+\sqrt5)/2$ then there are arbitrarily large $x$ such that $f_a(x)$ remains bounded away from zero).