The function $f:\mathbb{N} \rightarrow \mathbb{N}$ verifies:
$1) f(n+2)-2022 \cdot f(n+1)+2021 \cdot f(n)=0, \forall n \in \mathbb{N};$
$2) f(20^{22})=f(22^{20});$
$3) f(2021)=2022$.
Find all possible values of $f(2022)$.
1. **Define the sequence and the recurrence relation:**
   Let \( a_n \triangleq f(n) \). The given recurrence relation is:
   \[
   a_{n+2} - 2022 \cdot a_{n+1} + 2021 \cdot a_n = 0.
   \]

2. **Find the characteristic equation:**
   The characteristic equation associated with this recurrence relation is:
   \[
   \lambda^2 - 2022\lambda + 2021 = 0.
   \]

3. **Solve the characteristic equation:**
   Solve for \(\lambda\) using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   \lambda = \frac{2022 \pm \sqrt{2022^2 - 4 \cdot 2021}}{2}.
   \]
   Simplify the discriminant:
   \[
   2022^2 - 4 \cdot 2021 = 4088484 - 8084 = 4080400.
   \]
   Thus,
   \[
   \sqrt{4080400} = 2020.
   \]
   Therefore, the roots are:
   \[
   \lambda = \frac{2022 \pm 2020}{2}.
   \]
   This gives us:
   \[
   \lambda_1 = \frac{2022 + 2020}{2} = 2021, \quad \lambda_2 = \frac{2022 - 2020}{2} = 1.
   \]

4. **Form the general solution:**
   The general solution to the recurrence relation is:
   \[
   a_n = u \cdot 2021^n + v \cdot 1^n = u \cdot 2021^n + v.
   \]

5. **Apply the initial conditions:**
   - From the third condition \( f(2021) = 2022 \):
     \[
     a_{2021} = u \cdot 2021^{2021} + v = 2022.
     \]
   - From the second condition \( f(20^{22}) = f(22^{20}) \):
     \[
     u \cdot 2021^{20^{22}} + v = u \cdot 2021^{22^{20}} + v.
     \]
     Since \( 20^{22} \neq 22^{20} \), the only way this equation holds for all \( n \) is if \( u = 0 \).

6. **Determine the constant value:**
   With \( u = 0 \), the sequence simplifies to:
   \[
   a_n = v.
   \]
   Using the third condition again:
   \[
   v = 2022.
   \]

7. **Conclusion:**
   Therefore, the function \( f(n) \) is constant and equal to 2022 for all \( n \).

The final answer is \( \boxed{2022} \).