[tex]\sin x>[/tex] for [tex]0<x<\pi[/tex], but [tex]\sin\dfrac\pi n[/tex] is only decreasing for [tex]n\ge2[/tex], as you mentioned. This will not affect the behavior of the series. The sum starting from [tex]n=2[/tex] converges, so its value must be finite. Including the term generated by [tex]n=1[/tex] (which is finite, and zero besides, in this case) is the same as adding two finite numbers, which is yet another finite number. So the fact that the alternating series test only applies for a subset of the terms in the series doesn't matter.
For the second problem, the ratio test is an excellent choice. You have
[tex]\displaystyle\lim_{n\to\infty}\left|\dfrac{\frac{1\times3\times\cdots\times(2n-1)\times(2n+1)}{2\times5\times\cdots\times(3n-1)\times(3n+2)}}{\frac{1\times3\times\cdots\times(2n-1)}{2\times5\times\cdots\times(3n-1)}}\right|=\lim_{n\to\infty}\left|\frac{2n+1}{3n+2}\right|=\dfrac23<1[/tex]
which means the series converges.
