Five. (20 points) For what value of $a$ does the graph of the function $f(x)=(x+a)(|x+1-a|+|x-3|)-2x+4a$ have a center of symmetry?

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For the function $f(x)=(x+a)(|x+1-a|+|x-3|)-2x+4a$, on the right interval $x \geqslant \max \{3, a-1\}$, it takes the form
$$
\begin{array}{l}
f(x)=(x+a)(2x-a-2)-2x+4a \\
=2x^{2}+(a-4)x+q_{1} ;
\end{array}
$$

On the left interval $x \leqslant \min \{3, a-1\}$, it takes the form
$$
\begin{aligned}
f(x) & =(x+a)(-2x+a+2)-2x+4a . \\
& =-2x^{2}-ax+q_{2} ;
\end{aligned}
$$

In the middle segment, it is a linear function.
Therefore, if the graph of the function has a center of symmetry, it can only be the midpoint of this linear segment, with the x-coordinate $x_{0}=\frac{3+(a-1)}{2}=\frac{a+2}{2}$. The axes of symmetry of the parabolas on the left and right segments, $x=x_{1}$ and $x=x_{2}$, should also be symmetric about $x=x_{0}$, i.e., $x_{0}=\frac{x_{1} x_{2}}{2}=\frac{-\frac{a-4}{2}+\frac{a}{-4}}{2}=\frac{-a+2}{4}$. From these two conditions, we get $\frac{a+2}{2}=\frac{-a+2}{4}$. Thus,
$$
a=-\frac{2}{3}, x_{0}=\frac{2}{3} \text{. }
$$

Finally, it is not difficult to verify that $a=-\frac{2}{3}$ meets the requirements.
Indeed, when $a=-\frac{2}{3}$, the function $F(x)=f\left(x+x_{0}\right)-f\left(x_{0}\right)=x\left(\left|x+\frac{7}{3}\right|+\left|x-\frac{7}{3}\right|\right)-2x$ is an odd function, and its graph is symmetric about the origin.

Therefore, the function $f(x)$ is symmetric about the point $\left(x_{0}, f\left(x_{0}\right)\right)$, i.e., about the point $\left(\frac{2}{3},-4\right)$.