Three, (10 points) Let $a+b+c=0, a^{2}+b^{2}+c^{2}=1$.
(1) Find the value of $ab+bc+ca$;
(2) Find the value of $a^{4}+b^{4}+c^{4}$.
Three, (1) Since $a+b+c=0$, then,
$$
\begin{array}{l}
(a+b+c)^{2} \\
=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=0 .
\end{array}
$$

Given $a^{2}+b^{2}+c^{2}=1$, then $a b+b c+c a=-\frac{1}{2}$.
$$
\begin{array}{l}
\text { (2) } a b+b c+c a=-\frac{1}{2} \\
\Rightarrow(a b+b c+c a)^{2}=\frac{1}{4} \\
\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a b c(a+b+c)=\frac{1}{4} \\
\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=\frac{1}{4} .
\end{array}
$$

Given $a^{2}+b^{2}+c^{2}=1$, squaring it we get
$$
\begin{array}{l}
a^{4}+b^{4}+c^{4}+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}=1 \\
\Rightarrow a^{4}+b^{4}+c^{4}=1-2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) \\
\quad=1-2 \times \frac{1}{4}=\frac{1}{2} .
\end{array}
$$