11. (14 points) Given the function
$$
f(x)=2 \sin ^{2}\left(\frac{\pi}{4}-x\right)-\sqrt{3} \cos 2 x \text {. }
$$
(1) Find the smallest positive period and the intervals of monotonic decrease of $f(x)$;
(2) If $f(x)<m+2$ holds for all $x \in\left[0, \frac{\pi}{6}\right]$, find the range of real numbers $m$.
(1) Notice,
$$
\begin{array}{l}
f(x)=1-\cos \left(\frac{\pi}{2}-2 x\right)-\sqrt{3} \cos 2 x \\
=-(\sin 2 x+\sqrt{3} \cos 2 x)+1 \\
=-2 \sin \left(2 x+\frac{\pi}{3}\right)+1
\end{array}
$$

Thus, the smallest positive period of $f(x)$ is $T=\frac{2 \pi}{2}=\pi$.
From $2 k \pi-\frac{\pi}{2} \leqslant 2 x+\frac{\pi}{3} \leqslant 2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z})$
$$
\Rightarrow k \pi-\frac{5 \pi}{12} \leqslant x \leqslant k \pi+\frac{\pi}{12}(k \in \mathbf{Z}) \text {. }
$$

Therefore, the monotonic decreasing interval of $f(x)$ is
$$
\left[k \pi-\frac{5 \pi}{12}, k \pi+\frac{\pi}{12}\right](k \in \mathbf{Z}) \text {. }
$$
(2) Given $x \in\left[0, \frac{\pi}{6}\right]$, we have
$$
\begin{array}{l}
\frac{\pi}{3} \leqslant 2 x+\frac{\pi}{3} \leqslant \frac{2 \pi}{3} \\
\Rightarrow \frac{\sqrt{3}}{2} \leqslant \sin \left(2 x+\frac{\pi}{3}\right) \leqslant 1 .
\end{array}
$$

Thus, when $\sin \left(2 x+\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$, $f(x)$ attains its maximum value $1-\sqrt{3}$, i.e.,
$$
f(x)_{\max }=1-\sqrt{3} \text {. }
$$

To make $f(x)-1-\sqrt{3}$, the range of $m$ is $(-1-\sqrt{3},+\infty)$.