11. (20 points) Given the function
$$
f(x)=\frac{1}{3} x^{3}-(2 a+1) x^{2}+3 a(a+2) x+\frac{2}{3},
$$

where $a$ is a real number.
(1) When the graph of $y=f^{\prime}(x)$ has a unique common point with the $x$-axis in the interval $(0,6)$, find the range of the real number $a$;
(2) When $a=-1$, find the maximum and minimum values of the function $y=f(x)$ on the interval $[0,6]$.
11. Given
$$
f^{\prime}(x)=x^{2}-2(2 a+1) x+3 a(a+2) \text {. }
$$
(1) The x-coordinates of the intersection points of the graph of the function $y=f^{\prime}(x)$ with the x-axis are the real roots of the quadratic equation
$$
x^{2}-2(2 a+1) x+3 a(a+2)=0
$$

which are
$$
x_{1}=a+2, x_{2}=3 a \text {. }
$$

We will discuss the following three cases.
(i) When $x_{1}=x_{2}$, we have
$$
a+2=3 a \Rightarrow a=1 \text {. }
$$

Thus, $x_{1}=x_{2}=3$ is the only intersection point of the graph of the function $y=f^{\prime}(x)$ with the x-axis in the interval $(0,6)$, so $a=1$ is the solution.
(ii) When $x_{1}>x_{2}$, we have
$$
a+2>3 a \Rightarrow a<1 \text {. }
$$

Thus, $x_{2}<x_{1}<3$.
Since the graph of the function $y=f^{\prime}(x)$ has only one intersection point with the x-axis in the interval $(0,6)$, we have $3<x_{1}<6 \leqslant x_{2}$, which means
$$
3<a+2<6 \leqslant 3 a \text {. }
$$

Solving this, we get $2 \leqslant a<4$.
Therefore, the range of real values for $a$ is $-2<a \leqslant 0$, or $a=1$, or $2 \leqslant a<4$.
(2) When $a=-1$, we have
$$
f(x)=\frac{1}{3} x^{3}+x^{2}-3 x+\frac{2}{3} \text {, }
$$

and the roots of $f^{\prime}(x)=x^{2}+2 x-3=0$ are $x_{1}=1, x_{2}=-3$, with only $x_{1}=1$ in the interval $[0,6]$.

Thus, $f(x)$ is decreasing on $[0,1]$ and increasing on $[1,6]$, and
$$
f(0)=\frac{2}{3}, f(1)=-1, f(6)=90 \frac{2}{3} .
$$

Therefore, the maximum value of the function $y=f(x)$ on $[0,6]$ is $90 \frac{2}{3}$, and the minimum value is -1.