2. Given $x>0, y>0$, and satisfy
$$
\left\{\begin{array}{l}
\cos ^{2}(\pi x)+2 \sin (\pi y)=1, \\
\sin (\pi x)+\sin (\pi y)=0, \\
x^{2}-y^{2}=12 .
\end{array}\right.
$$

Then the ordered pair $(x, y)=$ $\qquad$ .
Solution: $(4,2)$.
From (1) and (2), we get $\sin (\pi x)[2+\sin (\pi x)]=0$.
Since $2+\sin (\pi x)>0$, it follows that $\sin (\pi x)=0$.
Substituting into (2) gives $\sin (\pi y)=0$.
Thus, $x$ and $y$ are both positive integers.
From (3), we get $(x-y)(x+y)=12=2^{2} \times 3$.
Since $x-y$ and $x+y$ have the same parity, and $x-y<x+y$, we have
$$
\left\{\begin{array}{l}
x - y = 2, \\
x + y = 6.
\end{array} \text{ Solving, we get } \left\{\begin{array}{l}
x=4, \\
y=2.
\end{array}\right.\right.
$$

Therefore, $(x, y)=(4,2)$.
$$
S_{\triangle A F_{1} B}=\frac{1}{2} \cdot 4 a \cdot 1=2 a .
$$

Also, $S_{\triangle A F_{1} B}=\frac{1}{2} \cdot 2 c \cdot\left|y_{1}-y_{2}\right|=4 c$.
Thus, from $2 a=4 c$, we get $e=\frac{c}{a}=\frac{1}{2}$.