The value of $a$ is chosen so that the number of roots of the first equation $4^{x}-4^{-x}=2 \cos a x, 4^{x}+4^{-x}=2 \cos a x+4$ is 2007.

How many roots does the second equation have for the same $a$?
Transform the second equation:

$$
\left.4^{x}+4^{-x}=2 \cos a x+4 \Leftrightarrow 4^{x}-2+4^{-x}=2(1+\cos a x) \Leftrightarrow\left(2^{x}-2^{-x}\right)^{2}=4 \cos ^{2 a x} / 2\right) \Leftrightarrow
$$

$$
\left[\begin{array} { l } 
{ 4 ^ { x / 2 } - 4 ^ { - x / 2 } = 2 \cos \frac { a x } { 2 } , } \\
{ 4 ^ { x / 2 } - 4 ^ { - x / 2 } = - 2 \cos \frac { a x } { 2 } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
4^{x / 2}-4^{-x / 2}=2 \cos \frac{a x}{2} \\
4^{-x / 2}-4^{x / 2}=2 \cos \frac{a x}{2}
\end{array}\right.\right.
$$

Both equations of this system can be reduced to the first equation in the problem's condition by the substitutions $x=2 y$ and $x=-2 z$ respectively. Therefore, each of these two equations has 2007 roots. If these equations have a common root $x=x_{0}$, then $4^{x_{0} / 2}-4^{-x_{0} / 2}=0$ and $\cos \frac{a x_{0}}{2}=0$, which is impossible. Therefore, these equations do not have common roots, and the second equation in the condition has $2 \cdot 2007=4014$ roots.

## Answer

4014.