Three, (25) Let $x_{1}, x_{2}, \cdots, x_{p}$ all be natural numbers, and $x_{1}<x_{2}<\cdots<x_{p}$, satisfying $2^{x_{1}}+2^{x_{2}}+\cdots+2^{x_{p}}$ $=2008$. Try to find the values of $p$ and $x_{1}, x_{2}, \cdots, x_{p}$.

Three, since $x_{1}, x_{2}, \cdots, x_{p}$ are all natural numbers, therefore, $2^{x_{1}}$, $2^{x_{2}}, \cdots, 2^{x_{p}}$ are all positive integers.
Also, since $x_{1}<x_{2}<\cdots<x_{p}$, it follows that
$$
2^{x_{1}}<2^{x_{2}}<\cdots<2^{x_{p}} \text {. }
$$

From $2^{x_{1}}+2^{x_{2}}+\cdots+2^{x_{p}}=2008$, we get
$$
2^{x_{1}}\left(1+2^{x_{2}-x_{1}}+2^{x_{3}-x_{1}}+\cdots+2^{x_{p}-x_{1}}\right)=2^{3} \times 251 \text {. }
$$

Since $2^{x_{1}}$ and $1+2^{x_{2}-x_{1}}+2^{x_{3}-x_{1}}+\cdots+2^{x_{p}-x_{1}}$ are coprime, and $2^{3}$ and 251 are also coprime, it must be that $2^{x_{1}}=2^{3}$, i.e., $x_{1}=3$; and $1+2^{x_{2}-x_{1}}+2^{x_{3}-x_{1}}+\cdots+2^{x_{p}-x_{1}}=251$. Thus,
$$
2^{x_{2}-x_{1}}+2^{x_{3}-x_{1}}+\cdots+2^{x_{p}-x_{1}}=250,
$$

which means $2^{x_{2}-x_{1}}\left(1+2^{x_{3}-x_{2}}+\cdots+2^{x_{p}-x_{2}}\right)=2 \times 125$.
$$
\begin{array}{l}
\text { Hence } x_{2}-x_{1}=1, x_{2}=x_{1}+1=4, \text { and } \\
1+2^{x_{3}-x_{2}}+\cdots+2^{x_{p}-x_{2}}=125 . \\
\cdots \cdots
\end{array}
$$

Continuing in this manner, using the same method, we can successively determine: $x_{3}=6$, $x_{4}=7, x_{5}=8, x_{6}=9, x_{7}=10$, and determine the value of $p$ to be 7.

Thus, $p=7, x_{1}=3, x_{2}=4, x_{3}=6, x_{4}=7, x_{5}=8, x_{6}=9, x_{7}=10$.