(2) Define $b-a$ as the "length" of the set $\{x \mid a \leqslant x \leqslant b\}$. Let $M=$ $\left\{x \left\lvert\, m \leqslant x \leqslant m+\frac{3}{4}\right.\right\}, N=\left\{x \left\lvert\, n-\frac{1}{3} \leqslant x \leqslant n\right.\right\}$, and both $M$ and $N$ are subsets of the set $\{x \mid 0 \leqslant x \leqslant 1\}$. Then the minimum "length" of the set $M \cap N$ is $\qquad$.
(2) $\frac{1}{12}$