Solution I We have
$$\begin{aligned} f(x+2) - f(x) &= -(f(x+4) - f(x+2)) - (f(x+6) - f(x+4)) + (f(x+6) - f(x)) \\ &\geq -3 \times 2^{x+2} - 3 \times 2^{x+4} + 63 \times 2^x = 3 \times 2^x. \end{aligned}$$
This means that $f(x+2) - f(x) = 3 \times 2^x$. So we have
$$\begin{aligned} f(2008) &= f(2008) - f(2006) + f(2006) - f(2004) + \dots \\ &\quad + f(2) - f(0) + f(0) \\ &= 3 \times (2^{2006} + 2^{2004} + \dots + 2^2 + 1) + f(0) \\ &= 3 \times \frac{4^{1003} + 1}{4 - 1} + 2008 \\ &= 2^{2008} + 2007. \end{aligned}$$

Solution II We define $g(x) = f(x) - 2^x$. Then we have
$$\begin{aligned} g(x+2) - g(x) &= f(x+2) - f(x) - 2^{x+2} + 2^x \\ &\leq 3 \times 2^x - 3 \times 2^x = 0, \\ g(x+6) - g(x) &= f(x+6) - f(x) - 2^{x+6} + 2^x \\ &\geq 63 \times 2^x - 63 \times 2^x = 0. \end{aligned}$$
This means that $g(x) \leq g(x+6) \leq g(x+4) \leq g(x+2) \leq g(x)$, and it implies that $g(x)$ is a periodic function with 2 as a period. So
$$\begin{aligned} f(2008) &= g(2008) + 2^{2008} = g(0) + 2^{2008} \\ &= 2007 + 2^{2008}. \end{aligned}$$