10. (20 points) Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms as $S_{n}$, and $S_{n}=2 n-a_{n}\left(n \in \mathbf{N}^{*}\right)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=2^{n-1} a_{n}$, prove that: $\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{n}}<\frac{5}{3}$. (Yang Zhiming provided the problem)
10. (1) (Method 1) It is easy to find that $a_{1}=1, a_{2}=\frac{3}{2}, a_{3}=\frac{7}{4}, a_{4}=\frac{15}{8}$. From this, we conjecture: $a_{n}=\frac{2^{n}-1}{2^{n-1}}$.

We will prove this by mathematical induction.
When $n=1$, $a_{1}=S_{1}=2-a_{1} \Rightarrow a_{1}=1$, the conclusion holds.
Assume that when $n=k\left(k \in \mathbf{N}^{*}\right)$, the conclusion holds, i.e., $a_{k}=\frac{2^{k}-1}{2^{k-1}}$, then when $n=k+1$, we have
$$
a_{k+1}=S_{k+1}-S_{k}=\left(2(k+1)-a_{k+1}\right)-\left(2 k-a_{k}\right)=2+a_{k}-a_{k+1},
$$

which gives $2 a_{k+1}=2+a_{k}$, thus
$$
a_{k+1}=\frac{2+a_{k}}{2}=\frac{2+\frac{2^{k}-1}{2^{k-1}}}{2}=\frac{2^{k+1}-1}{2},
$$

This shows that when $n=k+1$, the conclusion holds.
In summary, the general term formula for the sequence $\left\{a_{n}\right\}$ is $a_{n}=\frac{2^{n}-1}{2^{n-1}}$.
(Method 2) When $n=1$, $a_{1}=S_{1}=2-2 a_{1}$, solving gives $a_{1}=1$.
When $n \geqslant 2$, we have
$$
a_{n}=S_{n}-S_{n-1}=\left(2 n-a_{n}\right)-\left(2(n-1)-a_{n-1}\right)=2-a_{n}+a_{n-1} \text {, }
$$

which means
$$
a_{n}=\frac{1}{2} a_{n-1}+1,
$$

Multiplying both sides by $2^{n}$, we get
$$
2^{n} a_{n}=2^{n-1} a_{n-1}+2^{n},
$$

Thus
$$
2^{n} a_{n}=2 a_{1}+\left(2^{2}+2^{3}+\cdots+2^{n}\right)=2^{n+1}-2,
$$

Therefore, $a_{n}=\frac{2^{n}-1}{2^{n-1}}$. Upon verification, $a_{1}=1$ also fits this formula, so the general term formula for the sequence $\left\{a_{n}\right\}$ is $a_{n}=\frac{2^{n}-1}{2^{n-1}}$.
(Method 3) Similarly to Method 2, we get $a_{n}=\frac{1}{2} a_{n-1}+1$, thus
$$
a_{n}-2=\frac{1}{2}\left(a_{n-1}-2\right)
$$

Therefore, $\left\{a_{n}-2\right\}$ is a geometric sequence with the first term $a_{1}-2=-1$ and common ratio $\frac{1}{2}$, i.e., $a_{n}-2=-\frac{1}{2^{n-1}}$, thus the general term formula for the sequence $\left\{a_{n}\right\}$ is $a_{n}=2-\frac{1}{2^{n-1}}=\frac{2^{n}-1}{2^{n-1}}$.
(2) From (1), we know that $b_{n}=2^{n-1} a_{n}=2^{n-1} \cdot \frac{2^{n}-1}{2^{n-1}}=2^{n}-1$. Therefore, $\frac{1}{b_{n}}=\frac{1}{2^{n}-1}$. Thus,
$$
\begin{array}{l}
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{n}} \\
=\frac{1}{2-1}+\frac{1}{2^{2}-1}+\frac{1}{2^{3}-1}+\frac{1}{2^{4}-1}+\cdots+\frac{1}{2^{n}-1} \\
<\frac{1}{2-1}+\frac{1}{2^{2}-1}+\frac{1}{2^{3}-2}+\frac{1}{2^{4}-2^{2}}+\cdots+\frac{1}{2^{n}-2^{n-2}} \\
=\frac{1}{2-1}+\frac{1}{2^{2}-1}\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-2}}\right) \\
=\frac{1}{2-1}+\frac{1}{2^{2}-1} \cdot \frac{1-\frac{1}{2^{n-1}}}{1-\frac{1}{2}} \\
<\frac{1}{2-1}+\frac{1}{2^{2}-1} \cdot \frac{1}{1-\frac{1}{2}}=\frac{5}{3}
\end{array}
$$