One, (Full marks 30 points) Let $f(x)=x^{2}-(4 a-2) - x-6 a^{2}$ have the minimum value $m$ on the interval $[0,1]$. Try to write the expression for $m$ in terms of $a$, $m=F(a)$. And answer: For what value of $a$ does $m$ achieve its maximum value? What is this maximum value?

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$$
-, f(x)=[x-(2 a-1)]^{2}-10 a^{2}+4 a-1 .
$$
(i) When $2 a-1 \leqslant 0$, i.e., $a \leqslant \frac{1}{2}$, $f(x)$ is \. on $[0,1]$. At this time, $m=f(0)=-6 a^{2}$.
(ii) When $2 a-1 \geqslant 1$, i.e., $a \geqslant 1$, $f(x)$ is L. on $[0,1]$. At this time, $m=f(1)=-6 a^{2}-4 a+3$.
(iii) When $0<2 a-1<1$, i.e., $\frac{1}{2}<a<1$, the minimum value is obtained at the vertex of the parabola. At this time, $m=-10 a^{2}+4 a-1$.
Combining the above, we get the piecewise function for $m$
$$
m=F(a)=\left\{\begin{array}{l}
-6 a^{2}, \quad a \leqslant \frac{1}{2}, \\
-10 a^{2}+4 a-1, \frac{1}{2}<a<1, \\
-6 a^{2}-4 a+3, \quad a \geqslant 1 .
\end{array}\right.
$$

Next, consider finding the maximum value of $m$. We need to discuss $F(a)$ in segments.
When $a \leqslant \frac{1}{2}$, $m=-6 a^{2} \leqslant 0$, equality holds only when $a=0$, at which time the maximum value is 0.

When $-\frac{1}{2}<a<1$, $m=-10\left(a-\frac{1}{5}\right)^{2}-\frac{3}{5}$ $\leqslant-10\left(\frac{1}{2}-\frac{1}{5}\right)^{2}-\frac{3}{5}=-\frac{3}{2}$. At this time, the maximum value is $\frac{3}{2}$.

When $a \geqslant 1$, $m=-6\left(a+\frac{1}{3}\right)^{2}+\frac{11}{3}$ $\leqslant-6\left(1+\frac{1}{3}\right)^{2}+\frac{11}{3}=-7$. At this time, the maximum value is -7.

In summary, when $a=0$, $m$ achieves its maximum value of 0.