(1) $f^{\prime}(x)=-\sin x+m \sin x+m\left(x+\frac{\pi}{2}\right) \cos x=(m-1) \sin x+m\left(x+\frac{\pi}{2}\right) \cos x$, $f^{\prime \prime}(x)=(2 m-1) \cos x-m\left(x+\frac{\pi}{2}\right) \sin x$.
Since when $x \in(-\pi, 0)$, $\sin x < 0$ and $\cos x > 0$. Thus, if $m \leq 0$, then $(m-1) \sin x > 0$, $-m\left(x+\frac{\pi}{2}\right) \sin x > 0 \Rightarrow f^{\prime \prime}(x) > 0$; if $x \in\left(-\frac{\pi}{2}, 0\right)$, $(2 m-1) \cos x > 0$, $-m\left(x+\frac{\pi}{2}\right) \sin x > 0 \Rightarrow f^{\prime \prime}(x) > 0$. If $0 < m \leq 1$, then $f^{\prime}(0) = \frac{\pi}{2} m > 0$, combining the graph, we know that $f(x)$ has only one zero point $x = -\frac{\pi}{2}$ in the interval $(-\pi, 0)$; when $m = 0$, $f(x) = \cos x$, clearly $f(x)$ has only one zero point $x = -\frac{\pi}{2}$ in the interval $(-\pi, 0)$; when $0 < m \leq 1$, $f^{\prime \prime}(x) > 0 \Rightarrow f(x)$ has only one zero point $x = -\frac{\pi}{2}$ in the interval $(-\pi, 0)$.
In summary, when $m \leq 1$, $f(x)$ has only one zero point $x = -\frac{\pi}{2}$ in the interval $(-\pi, 0)$.
(2) Let $g(x) = 2x + \pi + |f(x)|$, then the problem is equivalent to the existence of $t > 0$, such that when $x \in \left(-\frac{\pi}{2} - t, -\frac{\pi}{2}\right)$, $g(x) < 0$ always holds. If $m > -1$, then $g^{\prime}(x) = 2 - f^{\prime}(x)$, $g^{\prime}\left(-\frac{\pi}{2}\right) = 2 - f^{\prime}\left(-\frac{\pi}{2}\right) = 1 + m > 0$, $g\left(-\frac{\pi}{2}\right) = 0$, contradiction; if $m = -1$, then $g^{\prime}(x_1) = 2 - f^{\prime}(x_1) = 2 > 0$, $g^{\prime}\left(-\frac{\pi}{2}\right) = 1 + m = 0 \Rightarrow g^{\prime}(x) \geq 0$, thus $g(x)$ is monotonically increasing on $\left(x_1, -\frac{\pi}{2}\right)$, when $x \in \left(x_1, -\frac{\pi}{2}\right)$, $g(x) < 0$, when $x \in \left(-\frac{\pi}{2} - t, -\frac{\pi}{2}\right)$, $|f(x)| < -2x - \pi$ always holds. Therefore, the minimum value of $m$ that satisfies the condition is -1.
14. Given the function $f(x)=\cos x+m\left(x+\frac{\pi}{2}\right) \sin x$, where $m \leqslant 1$.
(1) Discuss the number of zeros of $f(x)$ in the interval $(-\pi, 0)$;
(2) If there exists $t>0$, such that for $x \in\left(-\frac{\pi}{2}-t,-\frac{\pi}{2}\right)$, we always have $|f(x)|<-2 x-\pi$, find the minimum value of $m$ that satisfies this condition.