1. Given vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ satisfy:
$$
\begin{array}{l}
|a-b|=3,|a+2 b|=6, \\
a^{2}+a \cdot b-2 b^{2}=-9 .
\end{array}
$$

Then $|b|=$ $\qquad$
,$- 1 . \sqrt{7}$.
From the condition, we know
$$
\begin{array}{l}
a^{2}+a \cdot b-2 b^{2} \\
=(a-b) \cdot(a+2 b)=-9 .
\end{array}
$$

Then $|3 b|=|(a+2 b)-(a-b)|$
$$
\begin{array}{l}
=\sqrt{((a+2 b)-(a-b))^{2}} \\
=\sqrt{(a+2 b)^{2}+(a-b)^{2}-2(a+2 b) \cdot(a-b)} \\
=\sqrt{63}=3 \sqrt{7} .
\end{array}
$$

Therefore, $|\boldsymbol{b}|=\sqrt{7}$.