As you already observed, $F(h) = \int_{1/h}^\infty \frac{|\cos x|}{x^2}\,dx$. Now consider one of the pieces of the transformed integral between sign switches, on $[(n-1/2)\pi, (n+1/2)\pi]$. On this interval, $\frac{1}{(n+1/2)^2\pi^2} |\cos x| \le \frac{|\cos x|}{x^2} \le \frac{1}{(n-1/2)^2\pi^2} |\cos x|$. Therefore,
$$\frac{2}{(n+1/2)^2\pi^2} \le \int_{(n-1/2)\pi}^{(n+1/2)\pi} \frac{|\cos x|}{x^2}\,dx \le \frac{2}{(n-1/2)^2\pi^2}.$$
From this, we conclude:
$$\sum_{n=N}^\infty \frac{2}{(n+1/2)^2\pi^2} \le \int_{(N-1/2)\pi}^{\infty} \frac{|\cos x|}{x^2}\,dx \le \sum_{n=N}^\infty \frac{2}{(n-1/2)^2\pi^2}.$$
Now, since $\sum_{n=N}^\infty \frac{1}{(n+a)^2} \sim \frac{1}{N}$ as $N \to \infty$ for any constant $a$, we can conclude:
$$\int_{(N-1/2)\pi}^\infty \frac{|\cos x|}{x^2}\,dx \sim \frac{2}{N \pi^2}~\mathrm{as}~N\to \infty.$$
Now, for $(N-3/2)\pi \le X \le (N-1/2)\pi$, $\int_X^{(N-1/2)\pi} \frac{|\cos x|}{x^2}\,dx \le \frac{2}{(N-3/2)^2 \pi^2} = O(1/N^2)$. Therefore, in general,
$$\int_X^\infty \frac{|\cos x|}{x^2} \, dx \sim \frac{2}{X \pi}~\mathrm{as}~X \to \infty.$$
From this, we can conclude $\frac{F(h)}{h} \to \frac{2}{\pi}$ as $h \to 0^+$. Now, since $F$ is an odd function, $\frac{F(h)}{h}$ is even as a function of $h$, so in fact $\frac{F(h)}{h} \to \frac{2}{\pi}$ as $h \to 0$.
