Example 6 (2008 National High School Mathematics Joint Competition, First Test) 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 $f(2008)=$ $\qquad$ .
Solution 1 From the given conditions, we have
$$
\begin{aligned}
f(x+2)-f(x) & =-[f(x+4)-f(x+2)]-[f(x+6)-f(x+4)]+[f(x+6)-f(x)] \\
& \geqslant-3 \cdot 2^{x}-3 \cdot 2^{x+4}+63 \cdot 2^{x}=3 \cdot 2^{x},
\end{aligned}
$$

Therefore, $f(x+2)-f(x)=3 \cdot 2^{x}$, so
$$
\begin{aligned}
f(2008) & =f(2008)-f(2006)+f(2006)-f(2004)+\cdots+f(2)-f(0)+f(0) \\
& =3 \cdot\left(2^{2006}+2^{2004}+\cdots+2^{2}+1\right)+f(0) \\
& =3 \cdot \frac{4^{1004}-1}{4-1}+f(0) \\
& =2^{2008}+2007 .
\end{aligned}
$$

Solution 2 Let $g(x)=f(x)-2^{x}$, then
$$
\begin{array}{l}
g(x+2)-g(x)=f(x+2)-f(x)-2^{x+2}+2^{x} \leqslant 3 \cdot 2^{x}-3 \cdot 2^{x}=0, \\
g(x+6)-g(x)=f(x+6)-f(x)-2^{x+6}+2^{x} \geqslant 63 \cdot 2^{x}-63 \cdot 2^{x}=0,
\end{array}
$$

which means $g(x+2) \leqslant g(x) \cdot g(x+6) \geqslant g(x)$,
thus $g(x) \leqslant g(x+6) \leqslant g(x+4) \leqslant g(x+2) \leqslant g(x)$,
indicating that $g(x)$ is a periodic function with a period of 2,
so $f(2008)=g(2008)+2^{2008}=g(0)+2^{2008}=2^{2008}+2007$.