14. Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$.
14. (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$. From the problem, we know $a>0$, and $b+c=2-a, bc=\frac{4}{a}$. Therefore, $b, c$ are the two real roots of the quadratic equation $x^{2}-(2-a)x+\frac{4}{a}=0$, then
$$
\begin{array}{l}
\Delta=(2-a)^{2}-4 \times \frac{4}{a} \geqslant 0, \\
a^{3}-4 a^{2}+4 a-16 \geqslant 0, \\
\left(a^{2}+4\right)(a-4) \geqslant 0 .
\end{array}
$$

Thus, $a \geqslant 4$.
When $a=4, b=c=-1$, the conditions are satisfied.
Therefore, the minimum value of the maximum among $a, b, c$ is 4.
(2) Since $abc>0$, $a, b, c$ are either all positive or one positive and two negative.
(i) If $a, b, c$ are all positive, then from (1), the maximum of $a, b, c$ is not less than 4, which contradicts $a+b+c=2$.
(ii) If $a, b, c$ are one positive and two negative, let $a>0, b<0, c<0$, then
$$
\begin{array}{l}
|a|+|b|+|c|=a-b-c \\
=a-(2-a)=2a-2 .
\end{array}
$$

From (1), we know $a \geqslant 4$, hence $2a-2 \geqslant 6$.
When $a=4, b=c=-1$, the conditions are satisfied and the equality holds. Therefore, the minimum value of $|a|+|b|+|c|$ is 6.