For a Turing machine $M$, $\langle M \rangle$ denotes an encoding of $M$. Consider the following two languages. $$L_1 = \{\langle M \rangle \mid M \text{ takes more than } 2021 \text{ steps on all inputs}\}$$ $$L_2 = \{\langle M \rangle \mid M \text{ takes more than } 2021 \text{ steps on some input}\}$$ Which one of the following options is correct?

To determine whether $L_1$ and $L_2$ are decidable, consider how many tape cells a Turing machine $M$ can access in at most $2021$ steps:

1. In $2021$ computation steps, the tape head of $M$ can visit at most the first $2021$ tape cells starting from the initial position. Any input symbols located at positions beyond cell $2021$ cannot influence the machine's execution during the first $2021$ steps.

2. Since the input alphabet $\Sigma$ is finite, the set of all possible input prefixes of length at most $2021$ is finite (specifically, $\sum_{k=0}^{2021} |\Sigma|^k$).

3. Decidability of $L_1$: To check if $M$ takes more than $2021$ steps on *all* inputs, a decider can simulate $M$ for up to $2022$ steps on each of the finitely many input strings of length $\le 2021$. If $M$ halts within $2021$ steps on any such input, it rejects $\langle M \rangle$; otherwise, it accepts $\langle M \rangle$. Since the simulation always terminates in finite time, $L_1$ is decidable.

4. Decidability of $L_2$: Similarly, to check if $M$ takes more than $2021$ steps on *some* input, a decider simulates $M$ for $2022$ steps on each string of length $\le 2021$. If $M$ takes more than $2021$ steps on at least one string, it accepts; otherwise, it rejects. Thus, $L_2$ is also decidable.

Therefore, both $L_1$ and $L_2$ are decidable, which corresponds to Option (A).