1. Given the sequence of positive integers $\left\{a_{n}\right\}$ satisfies
$$
a_{n+2}=a_{n+1}+a_{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

If $a_{11}=157$, then $a_{1}=$ $\qquad$
-1.3 .
$$
\begin{array}{l}
\text { Given } a_{11}=a_{10}+a_{9}=2 a_{9}+a_{8} \\
=3 a_{8}+2 a_{7}=\cdots=55 a_{2}+34 a_{1},
\end{array}
$$

we know $55 a_{2}+34 a_{1}=157$.
Since $a_{1}, a_{2} \in \mathbf{Z}_{+}$, we have,
$$
55 \leqslant 55 a_{2} \leqslant 157 \text {. }
$$

Thus, $a_{2}=1$ or 2.
If $a_{2}=1$, then $a_{1}=3$;
If $a_{2}=2$, then $34 a_{1}=47$, which has no integer solution.
Therefore, $a_{1}=3$.