Problem 801

$11.9 \frac{1}{2}+4 \sqrt{5}$.
Let $f(x)=x^{2}-\left(a^{2}+b^{2}-6 b\right) x+a^{2}+b^{2}+2 a-4 b+1$.
From $f(x)=0$ with roots $x_{1} \leqslant 0 \leqslant x_{2} \leqslant 1$, we get $f(0) \leqslant 0, f(1) \geqslant 0$.
Simplifying, we obtain $(a+1)^{2}+(b-2)^{2} \leqslant 4$, and
$$a+b+1 \geqslant 0 \text {. }$$

In the coordinate system with $a$ and $b$ as the horizontal and vertical axes, respectively, draw the planning region represented by the above two inequalities. $a^{2}+b^{2}+4 a+4=(a+2)^{2}+b^{2}$ is the square of the distance from the point $(a, b)$ to the point $(-2,0)$.

Since the minimum value of the distance from the points in the planning region to the point $(-2,0)$ is the distance from the point $(-2,0)$ to the line $a+b+1=0$, which is $\frac{1}{\sqrt{2}}$, and the maximum distance from the points in the planning region to the point $(-2,0)$ is the sum of the distance from $(-2,0)$ to the center of the circle $(-1,2)$ and the radius 2, which is $\sqrt{5}+2$, the minimum value of $a^{2}+b^{2}+4 a+4$ is $\frac{1}{2}$; the maximum value is $(\sqrt{5}+2)^{2}=9+4 \sqrt{5}$.
Therefore, the sum of the minimum and maximum values is $9 \frac{1}{2}+4 \sqrt{5}$.