11. Let $f(x)$ be a function defined on $\mathbf{R}$. If $f(0)=2008$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+2)-f(x) \leqslant 3 \times 2^{x}, \\
f(x+6)-f(x) \geqslant 63 \times 2^{x},
\end{array}
$$

then $f(2008)=$
$$
11.2^{2008}+2007
$$

From the given conditions, we know
$$
\begin{array}{l} 
f(x+2)-f(x) \\
=-(f(x+4)-f(x+2))- \\
(f(x+6)-f(x+4))+ \\
(f(x+6)-f(x)) \\
\geqslant-3 \times 2^{x+2}-3 \times 2^{x+4}+63 \times 2^{x}=3 \times 2^{x} .
\end{array}
$$

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