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
$$
\begin{array}{l}
(a+c)(b+c)=a b+a c+b c+c^{2} \\
=a b+(a+b+c) c \\
=a b+\frac{1}{a b} \geqslant 2 \sqrt{a b \cdot \frac{1}{a b}}=2,
\end{array}
$$

the equality holds when $a b=1$, so when $a=b=1$, $c=\sqrt{2}-1$, $S$ can take the minimum value 2.
(2) When $S$ takes the minimum value, $a b=1$, thus,
$$
c(a+b+c)=1 \Rightarrow c^{2}+(a+b) c-1=0 \text {. }
$$

Let $t=a+b$. Then $t \geqslant 2 \sqrt{a b}=2$.
Thus, $c=\frac{-t+\sqrt{t^{2}+4}}{2}$ (negative value is discarded).
Since $c=\frac{-t+\sqrt{t^{2}+4}}{2}=\frac{2}{\sqrt{t^{2}+4}+t}$ is monotonically decreasing in $t \in$ $[2,+\infty)$, therefore, when $t=2$, $c$ has the maximum value $\sqrt{2}-1$