This is one of the well know examples of Weierstrass function. Hardy studied Hölder-continuity of such functions in Weierstrass's nondifferentiable function, G.H. Hardy, Trans. Amer. Math. Soc., 17 (1916), 301–325.. Here is a proof of Hölder-continuity for your case.
Theorem. Let $01$ and $ab>1$ then the function
$$
f(x)=\sum\limits_{n=1}^\infty a^n\cos(b^n x)
$$
is $(-\log_b a)$-Hölder continuous.
Proof.
Consider $x\in\mathbb{R}$ and $h\in(-1,1)$, then
$$
f(x+h)-f(x)=
\sum\limits_{n=1}^\infty a^{n}(\cos(b^n(x+h))-\cos(b^nx))=
$$
$$
-2\sum\limits_{n=1}^\infty a^{n}\sin(2^{-1}b^n(2x+h))\sin(2^{-1}b^{n}h)=
$$
Since $b>1$ and $h\in(-1,1)$ there exist $p\in\mathbb{N}$ such that $2^{-1}b^{p}|h|\leq 1
Corollary. For $\alpha\in(0,1)$ the function
$$
f(x)=\sum\limits_{n=1}^\infty 2^{-n\alpha}\cos(2^n x)
$$
is $\alpha$-Hölder continuous.
Proof Apply previous theorem with $a=2^{-\alpha}$ and $b=2$.
As for the proof of nowhere differentiability, I don't know a short proof. The problem is that the standard Weierstrass argument is not applicable here - parameters $a$, $b$ must satisfy inequality $ab>1+\frac{3\pi}{2}$. So it seems to me that one should repeat all the steps of Hardy's proof.