Three. (25 points) Given 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) 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 value of $a+b+c$.
(1) Given $a>0, c<0$, we know that $y^{\prime}=a x^{2}+b x+c$ intersects the $x$-axis, and $y_{\text {nuin }}^{\prime}<0$. Therefore,
$$
\left|y^{\prime}\right|=\left|a x^{2}+b x+c\right| \geqslant 0 \text {. }
$$

Thus, the minimum value of $\left|y^{\prime}\right|$ is 0.
At this point, $y_{\text {man }}=-2 \times 0-1=-1$.
(2) From $\left\{\begin{array}{l}y=a x^{2}+b x+c, \\ y=k(x-1)-\frac{k^{2}}{4} \text {, we get }\end{array}\right.$ $a x^{2}+(b-k) x+c+\frac{k^{2}}{4}+k=0$.
Let $\Delta=(b-k)^{2}-4 a\left(c+\frac{k^{2}}{4}+k\right)=0$.
Rearranging gives
$$
(1-a) k^{2}-(2 b+4 a) k+b^{2}-4 a c=0 \text {. }
$$

According to the problem,
$$
1-a=0,2 b+4 a=0, b^{2}-4 a c=0 \text {. }
$$

Solving these equations, we get $a=1, b=-2, c=1$.
Thus, $a+b+c=0$.