16. (15 points) Given the functions
$$
f(x)=a \sin x(a \in \mathbf{R}), g(x)=\mathrm{e}^{x} \text {. }
$$
(1) If $00, m0$ always holds, find the minimum value of $k$.
16. (1) Notice,
$F(x)=a \sin (1-x)+\ln x$,
$F^{\prime}(x)=-a \cos (1-x)+\frac{1}{x}$.
Since $0 < x < 1$,
we have $0 < 1-x < 1$.
Thus, $0 < \cos (1-x) < 1$.
Hence, $F^{\prime}(x) > 0$.
Therefore, $F(x)$ is monotonically increasing in the interval $(0,1)$.
(2) From (1), we know that when $a=1$,
$$
\begin{array}{l}
F(x) < 0 \text{ for } x \in (0,1) \text{ and } F(1) = 0.
\end{array}
$$
Let $u(x) = \mathrm{e}^x - 2mx - 2$,
$u^{\prime}(x) = \mathrm{e}^x - 2m$.
Since $m > 0$,
we have $u^{\prime}(x) > 0$ for $x > \ln (2m)$.
Thus, $u(x)$ is monotonically increasing in the interval $(0, +\infty)$.
Also, $u(0) = -1 < 0$ and $u(\ln 2) = 2 - 2m \ln 2 - 2 = -2m \ln 2 < 0$,
hence $u(x)$ has a unique zero $x_0$ in the interval $(0, \ln 2)$.
Then $\mathrm{e}^{x_0} - 2mx_0 - 2 = 0 \Rightarrow m = \frac{\mathrm{e}^{x_0} - 2}{2x_0}$.
Thus, when $x \in (0, x_0)$, $G^{\prime}(x) < 0$, $G(x)$ is monotonically decreasing,
and when $x \in (x_0, +\infty)$, $G^{\prime}(x) > 0$, $G(x)$ is monotonically increasing.
Therefore, $G(x)_{\min} = G(x_0)$
$= \mathrm{e}^{x_0} - mx_0^2 - 2(x_0 + 1) + k > 0$
$\Rightarrow k > -\mathrm{e}^{x_0} + \frac{\mathrm{e}^{x_0} - 2}{2x_0} x_0^2 + 2(x_0 + 1)$
$\Rightarrow k > \left(\frac{x_0}{2} - 1\right) \mathrm{e}^{x_0} + x_0 + 2$.
Let $v(x) = \left(\frac{x}{2} - 1\right) \mathrm{e}^x + x + 2$, then
$v^{\prime}(x) = \frac{x-1}{2} \mathrm{e}^x + 1$.
Let $t(x) = v^{\prime}(x) = \frac{x-1}{2} \mathrm{e}^x + 1$, then
$t^{\prime}(x) = \frac{x}{2} \mathrm{e}^x > 0$.
Thus, $v^{\prime}(x)$ is monotonically increasing in the interval $(0, \ln 2)$,
and $v^{\prime}(x) > v^{\prime}(0) = \frac{1}{2} > 0$.
Therefore, $v(x)$ is monotonically increasing in the interval $(0, \ln 2)$.
Hence, $1 = v(0) < v(x) < v(\ln 2) = 2 \ln 2 < 2$.
Thus, $k \geqslant 2$, and the minimum value of $k$ is 2.