One, (20 points) Let the function be
$$
f(x)=\cos 2 x+2 \sqrt{3} \sin x \cdot \cos x(x \in \mathbf{R})
$$

with the maximum value $M$ and the smallest positive period $T$.
(1) Find the values of $M$ and $T$, and write down the intervals where the function $f(x)$ is monotonically increasing;
(2) If 10 distinct positive numbers $x_{i}(i=1,2$, $\cdots, 10)$ satisfy $f\left(x_{i}\right)=M$, and $x_{i}<10 \pi$, find the value of $x_{1}+$ $x_{2}+\cdots+x_{10}$.
(1) Notice
$$
f(x)=2 \sin \left(2 x+\frac{\pi}{6}\right)(x \in \mathbf{R}) \text {. }
$$

Then $M=2, T=\frac{2 \pi}{2}=\pi$.
$$
\text { Also, } 2 k \pi-\frac{\pi}{2} \leqslant 2 x+\frac{\pi}{6} \leqslant 2 k \pi+\frac{\pi}{2}(k \in
$$
Z), then the monotonic increasing interval of the function $f(x)$ is
$$
\left[k \pi-\frac{\pi}{3}, k \pi+\frac{\pi}{6}\right](k \in \mathbf{Z}) .
$$
(2) From $f\left(x_{i}\right)=2 \sin \left(2 x_{i}+\frac{\pi}{6}\right)=2$, we get
$$
2 x_{i}+\frac{\pi}{6}=2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z}),
$$

which means $x_{i}=k \pi+\frac{\pi}{6}(k \in \mathbf{Z})$.
Since $0<x_{i}<10 \pi$, thus, $k=0,1, \cdots, 9$.
Therefore, $x_{1}+x_{2}+\cdots+x_{10}$
$$
=(0+1+\cdots+9) \pi+10 \times \frac{\pi}{6}=\frac{140 \pi}{3} .
$$