The first series converges, but the second one diverges.

Because $\sin{x} \le x$ holds for all $x\in \left[0, \frac{\pi}{2}\right]$ and the sine function is odd, therefore we have $|\sin{x}|\le |x|$ hold for all $x\in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Then, as the sine function is periodic, it can be shown that $|\sin{x}|\le |x|$ for all $x$. It follows that

$$0\le\sum_{n=1}^{\infty}\left|\sin\left(\frac{\cos{n}}{n^2}\right)\right|\le\sum_{n=1}^{\infty}\left|\frac{\cos{n}}{n^2}\right|\le\sum_{n=1}^{\infty}\left|\frac{1}{n^2}\right|$$

Now, because $\displaystyle\sum_{n=1}^{\infty}\left|\frac{1}{n^2}\right|$ is convergent, hence series $(A)$ converges absolutely and thus converges.

For the second series diverges because $$\lim_{n\to\infty}\frac{\sin{n}}{n^2}=0\implies\lim_{n\to\infty}\cos\left(\frac{\sin{n}}{n^2}\right)=1 > 0$$