3. Given that the equation with respect to $x$
$$
|1-| x||=m-2008
$$

has exactly two real solutions for $x$. Then the range of real values for $m$ is
3. $m=2008$ or $m>2009$.

Let $y=|1-| x||=m-2008$.
Sketch the graph of the function $y=$
$|1-| x||$ 
(see Figure 9).
When $y=0$ or $y$
$>1$, the line $y=$
$m-2008$ intersects the broken line
$y=|1-| x||$ at exactly two points, meaning there are exactly two real numbers $x$ that satisfy the condition.
Thus, $m-2008=0$ or $m-2008>1$.