How to prove these properties from the set Σ_m of null functions?

Question

A function $f : \mathbb{N} \to$ {0, 1} is called finally null if there exists $k \in \mathbb{N}$ such that, for all $n \geq k$, $f(n) = 0$. In this case, the rank of $f$, denoted by $\text{rk}(f)$, is defined as:$$\text{rk}(f) = \min\{ k \in \mathbb{N} : f(n) = 0, \forall n \geq k \}.$$For $m \in \mathbb{N}$, let $\Sigma_m$ be the set of all finally null functions $f : \mathbb{N} \to \{0, 1\}$ with rank $m$.

1. Prove that, for every $m \in \mathbb{N}$, the set $\Sigma_m$ is finite and calculate $\# \Sigma_m$ (the number of elements in $\Sigma_m$).
2. Prove that the set of all finally null functions $f : \mathbb{N} \to \{0, 1\}$ is countable.
3. Prove that the set $\{ A \subset \mathbb{N} : A \text{ is finite} \}$ is countable.

So, for question (1) I tried to solve it and deduced that its not null up to m-1, and null after m. Since each function can assume only 0 or 1, in the first m elements, the total number of functions is 2^m. I don't if I am correct, though.