3.2.29 ** Let the sequence of positive numbers $\left\{a_{n}\right\}$ satisfy: $a_{0}=1, a_{n}=a_{n+1}+a_{n+2}, n=0,1,2, \cdots$. Find $a_{1}$.
From the given, we have $a_{n+2}=-a_{n+1}+a_{n}$. Therefore,
$$
\begin{aligned}
a_{n} & =-a_{n-1}+a_{n-2}=-\left(-a_{n-2}+a_{n-3}\right)+a_{n-2} \\
& =2 a_{n-2}-a_{n-3}=2\left(-a_{n-3}+a_{n-1}\right)-a_{n-3} \\
& =-3 a_{n-3}+2 a_{n-4}=\cdots=(-1)^{n-1}\left(f_{n-1} a_{1}-f_{n-2}\right),
\end{aligned}
$$

where $\left\{f_{n}\right\}$ is the Fibonacci sequence (it is easy to prove the above formula using mathematical induction).
To ensure $a_{n}>0\left(n \in \mathbf{N}^{\star}\right)$, it is necessary that $a_{1}>\frac{f_{n-2}}{f_{n-1}}(n>1, n \text { is odd }) \text {. }
$$

By the general term formula of $f_{n}$ and $\left(\frac{\sqrt{5}-1}{2}\right)^{n} \rightarrow 0(n \rightarrow+\infty)$, we get $\frac{f_{n-1}}{f_{n-2}} \rightarrow \frac{1+\sqrt{5}}{2}$. Thus, from (1) and (2), we know that
$a_{1}=\frac{2}{1+\sqrt{5}}=\frac{\sqrt{5}-1}{2}$. Conversely, when $a_{1}=\frac{\sqrt{5}-1}{2}$, both (1) and (2) hold.
Therefore, $a_{1}=\frac{\sqrt{5}-1}{2}$.