11 Let $f(x)$ be a function defined on $\mathbf{R}$, if $f(0)=2008$, and for any $x \in \mathbf{R}$, it satisfies $f(x+2)-f(x) \leqslant 3 \cdot 2^{x}, f(x+6)-f(x) \geqslant 63 \cdot 2^{x}$, then $f(2008)=$ $\qquad$ .
11 Method 1: From the given conditions, we have
$$
\begin{aligned}
f(x+2)-f(x)= & -(f(x+4)-f(x+2))-(f(x+6) \\
& -f(x+4))+(f(x+6)-f(x)) \\
\geqslant & -3 \cdot 2^{x+2}-3 \cdot 2^{x+4}+63 \cdot 2^{x} \\
= & 3 \cdot 2^{x},
\end{aligned}
$$

Therefore, $f(x+2)-f(x)=3 \cdot 2^{x}$, so
$$
\begin{aligned}
f(2008)= & f(2008)-f(2006)+f(2006)-f(2004)+\cdots \\
& +f(2)-f(0)+f(0) \\
= & 3 \cdot\left(2^{2006}+2^{2004}+\cdots+2^{2}+1\right)+f(0) \\
= & 3 \cdot \frac{4^{1003+1}-1}{4-1}+f(0) \\
= & 2^{2008}+2007 .
\end{aligned}
$$

Method 2: Let $g(x)=f(x)-2^{x}$, then
$$
\begin{aligned}
g(x+2)-g(x) & =f(x+2)-f(x)-2^{x+2}+2^{x} \\
& \leqslant 3 \cdot 2^{x}-3 \cdot 2^{x} \\
& =0, \\
g(x+6)-g(x) & =f(x+6)-f(x)-2^{x+6}+2^{x} \\
& \geqslant 63 \cdot 2^{x}-63 \cdot 2^{x} \\
& =0,
\end{aligned}
$$

which means $g(x+2) \leqslant g(x), g(x+6) \geqslant g(x)$, hence
$$
g(x) \leqslant g(x+6) \leqslant g(x+4) \leqslant g(x+2) \leqslant g(x),
$$

so $g(x)$ is a periodic function with a period of 2, thus
$$
\begin{aligned}
f(2008) & =g(2008)+2^{2008} \\
& =g(0)+2^{2008} \\
& =2^{2008}+2007 .
\end{aligned}
$$