a)
We have
$$\begin{align*} f(x + \pi) &= a(|\sin(x + \pi)| + |\cos(x + \pi)|) - 3\sin(2x + 2\pi) - 7 \\ &= a(|-\sin x| + |-\cos x|) - 3\sin(2x) - 7 = f(x), \\ \\ f\left(\frac{\pi}{2} - x\right) &= a\left(|\sin\left(\frac{\pi}{2} - x\right)| + |\cos\left(\frac{\pi}{2} - x\right)|\right) - 3\sin(\pi - 2x) - 7 \\ &= a\left(|\cos x| + |\sin x|\right) - 3\sin(2x) - 7 = f(x), \\ \\ f\left(\frac{3\pi}{2} - x\right) &= a\left(|\sin\left(\frac{3\pi}{2} - x\right)| + |\cos\left(\frac{3\pi}{2} - x\right)|\right) - 3\sin(3\pi - 2x) - 7 \\ &= a\left(|-\cos x| + |-\sin x|\right) - 3\sin(2x) - 7 = f(x). \end{align*}$$

b)
Direct verification shows that for every integer $k$ we have $f(\frac{k\pi}{2}) = a-7$.
Moreover $f(\frac{\pi}{4}) = a\sqrt{2} - 10$ and $f(\frac{3\pi}{4}) = a\sqrt{2} - 4$. If $a \neq 7$, $a \neq 5\sqrt{2}$ and $a \neq 2\sqrt{2}$, according to a) the equation $f(x) = 0$ has even number of roots in any of the intervals $(0, \frac{\pi}{2})$ $(\frac{\pi}{2}, \pi)$ and therefore it has even number of roots in the interval $(0, n\pi)$.

1. If $a = 7$ then $f(x) = 7(|\sin x| + |\cos x|) - 3\sin 2x - 7$ and $f(\frac{\pi}{2}) = 0$.

1.1. Let $x \in (0, \frac{\pi}{2})$. Then $f(x) = 7(\sin x + \cos x) - 3\sin 2x - 7$. Set $y = \sin x + \cos x$. Then $y = \sqrt{2}\sin(x + \frac{\pi}{4}) \in (1, \sqrt{2}]$ ($\sin 2x = y^2 - 1$) and the equation $f(x) = 0$ becomes $3y^2 - 7y + 4 = 0$. Therefore $y_1 = 1$ and $y_2 = \frac{4}{3}$. Hence $y_2 = \frac{4}{3}$ and the equation $f(x) = 0$ has 2 roots in $(0, \frac{\pi}{2})$.

1.2. Let $x \in (\frac{\pi}{2}, \pi)$. Then $f(x) = 7(\sin x - \cos x) - 3\sin 2x - 7$. Set $y = \sin x - \cos x$. Then $y = \sqrt{2}\sin(x + \frac{\pi}{4}) \in (1, \sqrt{2}]$ and $f(x) = 0$ is equivalent to $3y^2 + 7y - 10 = 0$. Therefore $y_1 = 1$ and $y_2 = -\frac{10}{3}$, i.e. $f(x) = 0$ has no solutions in the given interval.
Therefore $f(x) = 0$ has 3 roots in the interval $(0, \pi)$. The total number of roots in the interval $(0, n\pi)$ equals $3n + n - 1 = 4n - 1$ and $4n - 1 = 2007$ implies $n = 502$.

2. If $a = 5\sqrt{2}$ then $f(x) = 5\sqrt{2}(|\sin x| + |\cos x|) - 3\sin 2x - 7$.

2.1. Let $x \in (0, \frac{\pi}{2})$. Then $f(x) = 5\sqrt{2}(\sin x + \cos x) - 3\sin 2x - 7$ and setting $y = \sin x + \cos x$ yields that $f(x) = 0$ is equivalent to $3y^2 - 5\sqrt{2}y + 4 = 0$, $y \in (1, \sqrt{2}]$. Therefore $y_1 = \sqrt{2}$ and $y_2 = \frac{2\sqrt{2}}{3} < 1$, i.e. only $x = \frac{\pi}{4}$ is a solution.

2.2. Let $x \in (\frac{\pi}{2}, \pi)$. Then $f(x) = 5\sqrt{2}(\sin x - \cos x) - 3\sin 2x - 7$ and setting $y = \sin x - \cos x$ yields that $f(x) = 0$ is equivalent to $3y^2 + 5\sqrt{2}y - 10 = 0$, $y \in (1, \sqrt{2}]$. The roots of this equation do not belong to the interval $(1, \sqrt{2}]$. In this case there is 1 root in $(0, \pi)$ and $n$ roots in $(0, n\pi)$. Therefore $n = 2007$.

3. When $a = 2\sqrt{2}$ analogous observations show that there is a unique root $x = \frac{3\pi}{4}$ in the interval $(0, \pi)$ and $n$ roots in the interval $(0, n\pi)$. Again $n = 2007$.

Answer: $a = 7$, $n = 502$; $a = 5\sqrt{2}$, $n = 2007$; $a = 2\sqrt{2}$, $n = 2007$.