13. Given the functions $f(x)=a x(a \in \mathbf{R}), g(x)=\ln x$.
(1) If the function $F(x)=f(x)-g(x)$ has an extremum of 1, find the value of $a$;
(2) If the function $G(x)=f(\sin (1-x))+g(x)$ is increasing on the interval $(0,1)$, find the range of values for $a$;
(3) Prove: $\sum_{k=1}^{n} \sin \frac{1}{(k+1)^{2}}<\ln 2$.
13. (1) From the problem, we know
$$
F(x)=a x-\ln x(x>0) \text {. }
$$

Thus, $F^{\prime}(x)=a-\frac{1}{x}(x>0)$.
(i) When $a \leqslant 0$, $F(x)$ is monotonically decreasing on $(0,+\infty)$, with no extremum.
(ii) When $a>0$, $F^{\prime}(x)=0 \Rightarrow x=\frac{1}{a}$. Therefore, $F(x)$ is monotonically decreasing on $\left(0, \frac{1}{a}\right)$ and monotonically increasing on $\left(\frac{1}{a},+\infty\right)$.
Thus, $F(x)$ has a minimum value at $x=\frac{1}{a}$, i.e.,
$$
F\left(\frac{1}{a}\right)=1-\ln \frac{1}{a} \Rightarrow 1-\ln \frac{1}{a}=1 \Rightarrow a=1 .
$$
(2) Note that, $G(x)=a \sin (1-x)+\ln x$, and it is an increasing function on the interval $(0,1)$.
Thus, for $x \in(0,1)$, we have
$$
G^{\prime}(x)=-a \cos (1-x)+\frac{1}{x} \geqslant 0 \text {. }
$$

Since $x \in(0,1)$, we have $\cos (1-x)>0$.
When $a \leqslant 0$, it is clear that
$$
G^{\prime}(x)=-a \cos (1-x)+\frac{1}{x} \geqslant 0 \text {. }
$$

When $a>0$,
$$
\begin{array}{l}
G^{\prime}(x)=-a \cos (1-x)+\frac{1}{x} \geqslant 0 \\
\Leftrightarrow \frac{1}{a} \geqslant x \cos (1-x) .
\end{array}
$$

Let $h(x)=x \cos (1-x)$.
Clearly, $h(x)=x \cos (1-x)$ is monotonically increasing on $x \in(0,1)$.
Thus, $h(x)<h(1)=1$.
From $\frac{1}{a} \geqslant 1 \Rightarrow 0<a \leqslant 1$.
In summary, $a \in(-\infty, 1]$.
(3) From (2), we know that when $a=1$,
$$
G(x)=\sin (1-x)+\ln x
$$

is an increasing function on the interval $(0,1)$.
Therefore, when $x \in(0,1)$,
$$
\begin{aligned}
G(x) & =\sin (1-x)+\ln x<G(1)=0 \\
\Rightarrow \sin (1-x) & <\ln \frac{1}{x} .
\end{aligned}
$$

Let $1-x=t$. Then when $t \in(0,1)$, $\sin t<\ln \frac{1}{1-t}$.

For $k \in \mathbf{Z}_{+}$, we have $\frac{1}{(k+1)^{2}} \in(0,1)$, thus,
$$
\begin{array}{l}
\sin \frac{1}{(k+1)^{2}}<\ln \frac{1}{1-\frac{1}{(k+1)^{2}}}=\ln \frac{(k+1)^{2}}{k(k+2)} . \\
\text { Hence } \sum_{k=1}^{n} \sin \frac{1}{(k+1)^{2}}<\sum_{k=1}^{n} \ln \frac{(k+1)^{2}}{k(k+2)} \\
=\ln \frac{2(n+1)}{n+2}<\ln 2 .
\end{array}
$$