I need to know the largest open set on which Laurent series of $f(z)={z^2-\pi^2\over \sin z}$ converges.
$f(z)={(z-\pi)(z+\pi)\over \sin z}$ clearly as $z\to \pm\pi,f(z)\to \pm2\pi$, so $f(z)$ extend to be analytic at $z=\pm\pi$
Can I say directly from this observation that the largest open set is $\{z:0<|z|<2\pi\}$?
Yes, since $\pm \pi$ are removable singularities of $f$, the closest poles to $0$ are $\pm 2\pi$, so the Laurent series converges in the punctured disk $\{ 0 < \lvert z\rvert < 2\pi\}$.
A Laurent series always converges in the largest annulus on which the function is holomorphic (after removing any removable singularities).