9. The last three digits of $9^{2022}$ are

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Since $(9,125)=1, \varphi(125)=125 \times \frac{4}{5}=100$, by Euler's theorem we get $9^{100} \equiv 1(\bmod 125)$.
Then $9^{2022} \equiv 9^{22}(\bmod 125)$, and $9^{3}=729 \equiv-21 \Rightarrow 9^{9} \equiv 441 \times(-21)$
$\equiv 66 \times(-21) \equiv-11(\bmod 125)$, thus $9^{18} \equiv 121 \equiv-4(\bmod 125)$,
$9^{22} \equiv(-4) \times(-21) \times 9=756 \equiv 6(\bmod 125)$. Therefore, the last three digits of $9^{22}$ could be $006,131,256,381,506$,
$631,756,881$, combining with $9^{2022} \equiv 1(\bmod 8)$ we get the last three digits of $9^{2022}$ are 881.