3. Solution. Since $\left(1-\frac{1}{k^{2}}\right)=\left(1-\frac{1}{k}\right)\left(1+\frac{1}{k}\right)=\frac{k-1}{k} \cdot \frac{k+1}{k}$, then

$$
\begin{aligned}
& \left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \cdots\left(1-\frac{1}{2021^{2}}\right)\left(1-\frac{1}{2022^{2}}\right)= \\
& \quad=\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{2}{3} \cdot \cdot \frac{4}{3} \cdot \frac{3}{4} \cdot \cdot \frac{5}{4} \cdots \frac{2020}{2021} \cdot \frac{2022}{2021} \cdot \frac{2021}{2022} \cdot \frac{2023}{2022}
\end{aligned}
$$

It is easy to notice that all fractions appear in pairs: $\frac{k+1}{k} \cdot \frac{k}{k+1}$. Each pair in the product gives 1. Thus, the desired product is $\frac{1}{2} \cdot \frac{2023}{2022}=\frac{2023}{4044}$.
Answer: $\frac{2023}{4044}$.