2. Given $x>0, y>0$, and satisfying $\left\{\begin{array}{l}\cos ^{2}(\pi x)+2 \sin (\pi y)=1 \\ \sin (\pi x)+\sin (\pi y)=0 \\ x^{2}-y^{2}=12\end{array}\right.$, then the ordered pair $(x, y)=$
From (1) and (2), we get $2 \sin (\pi y)=1-\cos ^{2}(\pi x)=\sin ^{2}(\pi x)=\sin ^{2}(\pi y)$ $\Rightarrow \sin (\pi x)=\sin (\pi y)=0 \Rightarrow x, y \in \mathbf{Z}$. Combining with the conditions, let $y=x-k(k>0)$, $x^{2}-y^{2}=x^{2}-(x-k)^{2}=2 k x-k^{2}=k(2 x-k)=12$. Thus, $k, 2 x-k$ are both even $\Rightarrow\left\{\begin{array}{l}k=2, \\ 2 x-k=6\end{array}\right.$ or $\left\{\begin{array}{l}k=6, \\ 2 x-k=2\end{array} \Rightarrow\left\{\begin{array}{l}x=4, \\ y=2\end{array}\right.\right.$. Therefore, the ordered pair $(x, y)=(4,2)$ satisfies the conditions.