3. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{2}=2, a_{2 k+1}=\frac{a_{2 k}^{2}}{a_{2 k-1}}$, and $a_{2 k+2}=2 a_{2 k+1}-a_{2 k}\left(k \in \mathbf{N}^{*}\right)$, then the last two digits of $a_{2022}$ are
$a_{2 k-1}, a_{2 k}, a_{2 k+1}$ form a geometric sequence, $a_{2 k}, a_{2 k+1}, a_{2 k+2}$ form an arithmetic sequence, the first few terms of the sequence $\left\{a_{n}\right\}$ are: $1,2,4,6,9,12,16$, $20,25,30,36, \cdots$, it is easy to prove by mathematical induction that $a_{2 n-1}=n^{2}, a_{2 n}=n(n+1)$.
Therefore, $a_{2022}=1011 \cdot 1012 \equiv 11 \cdot 12 \equiv 32(\bmod 100)$.