1. As shown in Figure 1, in square $A B C D$, points $M$ and $N$ are on sides $A D$ and $C D$ respectively, and $M D=2 A M, N D=3 C N$. Then $\sin \angle M B N=$
2.1. $\frac{11 \sqrt{170}}{170}$.
Connect $M N$. Without loss of generality, let $A D=12$. Then
$$
A M=4, D M=8, C N=3, D N=9 \text {. }
$$

Thus, $B M=\sqrt{A B^{2}+A M^{2}}=4 \sqrt{10}$,
$$
\begin{array}{l}
B N=\sqrt{B C^{2}+C N^{2}}=3 \sqrt{17} . \\
\text { Also, } S_{\triangle B B M}=24, S_{\triangle B C N}=18, S_{\triangle O M N}=36,
\end{array}
$$
$S_{\text {square } A B C D}=144$, hence,
$$
\begin{array}{l}
S_{\triangle B M N}=S_{\text {square } A B C D}-S_{\triangle A B M}-S_{\triangle B C N}-S_{\triangle D M N} \\
=144-24-18-36=66 .
\end{array}
$$

Therefore, $S_{\triangle B M N}=\frac{1}{2} B M \cdot B N \sin \angle M B N$
$$
\begin{array}{l}
\Rightarrow 66=\frac{1}{2} \times 4 \sqrt{10} \times 3 \sqrt{17} \sin \angle M B N \\
\Rightarrow \sin \angle M B N=\frac{11}{\sqrt{170}}=\frac{11 \sqrt{170}}{170} .
\end{array}
$$