(14) Given $a, b, c \in \mathbf{R}^{+}$, satisfying $a b c(a+b+c)=1$.
(1) Find the minimum value of $S=(a+c)(b+c)$;
(2) When $S$ takes the minimum value, find the maximum value of $c$.

保留了源文本的换行和格式。
14 （1）Since $(a+c)(b+c)=a b+a c+b c+c^{2}$
$$
\begin{array}{l}
=a b+(a+b+c) c \\
=a b+\frac{1}{a b} \geqslant 2 \sqrt{a b \cdot \frac{1}{a b}}=2,
\end{array}
$$

The equality holds when $a b=1$, and when $a=b=1, c=\sqrt{2}-1$, $S$ can achieve the minimum value of 2.
（2）When $S$ takes the minimum value, $a b=1$, thus $c(a+b+c)=1$, i.e., $c^{2}+$ $(a+b) c-1=0$. Let $t=a+b$, then $t \geqslant 2 \sqrt{a b}=2$.
Therefore, $c=\frac{-t+\sqrt{t^{2}+4}}{2}$ or $c=\frac{-t-\sqrt{t^{2}+4}}{2}<0$ (discard). Hence, $c=\frac{-t+\sqrt{t^{2}+4}}{2}=\frac{2}{\sqrt{t^{2}+4}+t}$ is monotonically decreasing in $t \in[2,+\infty)$, so at $t=2$, $c$ has the maximum value $\sqrt{2}-1$.