(20 points)(1) Draw the graph of the function $f(x)=|3| x|-1|$;
(2) If the equation $|3| x|-1|=2^{x}+$ $a$ has 4 real roots with respect to $x$, find the set of values for the real number $a$.
(1) As shown in Figure 2.
(2) Let
$$
y=|3| x|-1| \text{, }
$$

then $y=2^{x}+a$.
When the graph of the function $y=2^{x}+a$ passes through the point $A(0,1)$, we have $1=2^{0}+a$, which gives $a=0$. At this point, the graph of $y=2^{x}$ intersects the ray $BC$ at one point $A'$, and intersects the ray $DE$ at two points $A''$ and $A'''$. From the equation $2^{x}=3x-1$, we get $A''(1,2)$ and $A'''(3,8)$.

When the graph of the function $y=2^{x}+a$ passes through the point $B\left(-\frac{1}{3}, 0\right)$, we have $0=2^{-\frac{1}{3}}+a$, which gives $a=-2^{-\frac{1}{3}}$. At this point, the graph of the function $y=2^{x}-2^{-\frac{1}{3}}$ intersects the line segment $DA$ at one point $B'$, and intersects the ray $DE$ at two points $B''$ and $B'''$.

Additionally, when $a>0$ or $a<-2^{-\frac{1}{3}}$, there are at most 3 intersection points; when $-2^{-\frac{1}{3}}<a<0$, there are 5 intersection points.
Therefore, the set of values for the real number $a$ is $\left\{0,-2^{-\frac{1}{3}}\right\}$.