Three. (Full marks 14 points) For the quadratic trinomial $a x^{2}+b x+c$ $(a>0)$.
(1) When $c<0$, find the maximum value of the function $y=-2\left|a x^{2}+b x+c\right|-1$;
(2) If for any real number $k$, the line $y=k(x-1)-$ $\frac{k^{2}}{4}$ intersects the parabola $y=a x^{2}+b x+c$ at exactly one point, find the values of $a, b, c$.
(1) $\because a>0, c>0$, hence the minimum value of $\left|a x^{2}+b x+c\right|$ is 0.
Then the maximum value of $y=-2\left|a x^{2}+b x+c\right|-1$ is -1.
(2) To make the line $y=k(x-1)-\frac{k^{2}}{4}$ intersect the parabola $y=a x^{2}+b x+c$ at only one point, the system of equations
$$
\left\{\begin{array}{l}
y=k(x-1)-\frac{k^{2}}{4}, \\
y=a x^{2}+b x+c
\end{array}\right.
$$
$$
a x^{2}+(b-k) x+\frac{k^{2}}{4}+k+c=0 .
$$
$\because$ This equation has only one solution, $\therefore \Delta=0$.
Rearranging gives the quadratic equation in $k$
$$
(1-a) k^{2}-2(2 a+b) k+b^{2}-4 a c=0 \text {. }
$$

$\because$ This equation holds for any real number $k$,
$\therefore$ all coefficients are 0.
Thus, $a=1, b=-2, c=1$.