18. (1) In $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$, replace $a, b, c$ with $\frac{a}{a+b+c}, \frac{b}{a+b+c}$, and $\frac{c}{a+b+c}$ respectively, and assume without loss of generality that $a+b+c=1$,
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}$$

Consider $f(x)=\frac{x}{1-x}$, where $0<x<1$. The equality in the inequality holds if and only if $a=b=c=\frac{1}{3}$. The tangent line equation of $y=f(x)$ at $x=\frac{1}{3}$ is $y=\frac{9 x-1}{4}$. Direct calculation yields
$$f(x)-\frac{9 x-1}{4}=\frac{(3 x-1)^{2}}{1-x} \geqslant 0$$

Therefore,
$$\begin{array}{l} \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=f(a)+f(b)+f(c) \geqslant \\ \frac{9 a-1}{4}+\frac{9 b-1}{4}+\frac{9 c-1}{4}= \\ \frac{9(a+b+c)-3}{4}=\frac{9-3}{4}=\frac{3}{2} \end{array}$$
(2) To prove $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2}$, it suffices to prove $\frac{a^{2}}{(b+c)(a+b+c)}+\frac{b^{2}}{(c+a)(a+b+c)}+\frac{c^{2}}{(a+b)(a+b+c)} \geqslant \frac{1}{2}$. In the left-hand side expression, replace $a, b, c$ with $\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}$ respectively, and assume without loss of generality that $a+b+c=F$. Consider $f(x)=\frac{x^{2}}{\models x}$, where $\theta<x<F$. The equality in the inequality holds if and only if $a=b=c=\frac{1}{3}$. The tangent line equation of $y=f(x)$ at $x=\frac{1}{3}$ is $y=\frac{5 x-1}{4}$. Direct calculation yields $f(x)-\frac{5 x-1}{4}=\frac{(3 x-1)^{2}}{1-x} \geqslant 0$, so,
$$\begin{array}{l} \frac{a^{2}}{1-a}+\frac{b^{2}}{1-b}+\frac{c^{2}}{1-c}=f(a)+f(b)+f(c) \geqslant \\ \frac{5 a-1}{4}+\frac{5 b-1}{4}+\frac{5 c-1}{4}= \\ \frac{5(a+b+c)-3}{4}=\frac{5-3}{4}=\frac{1}{2} \end{array}$$