$3$ is not a primitive root $\mod 100$. (Most residues are not.) So we don't need $3^{40}\equiv 1$. $3^2 = (10 - 1); 3^{20} = (10-1)^{10}\equiv -10*10 +1\equiv 1 \mod 100$. So $3^{20} \equiv 1 \mod 100$ will do.
So $a_2=3^3 = 27\equiv 7 \mod 20$
$a_3 = 3^{27}\equiv 3^7= (10-1)^3*3 \equiv (30 - 1)*3 \equiv 87\mod 100$ and $a_3 \equiv 7 \mod 20$.
$a_4 = 3^{a_3}\equiv 3^7\equiv 87 \mod 100$.
So $a_n \equiv 87 \mod 100$ for all $n \ge 3$.
As you've noticed, since $3\mid a_1$ and $3\mid 1983$, it follows that $3\mid a_n$ for all $n$. This allows us to simplify the problem by considering the associated sequence defined by $b_n = a_n/3$. We can easily prove by induction that we have $1 \le b_n \le 660$ for all $n$. Since the admissible range of values for $b_n$ is finite, the sequence must be eventually periodic.
Your conjecture that the period is $660$ is in fact true. Showing that the period is $660$ will show that the sequence is not just eventually periodic, but fully periodic (alternatively, as you've noted, this follows from the fact that $b_n$ uniquely determines $b_{n-1}$). The conjecture that the period is $660$, together with the fact that $1 \le b_n \le 660$, motivates looking at the values of the sequence modulo $661$.
Let $[k]$ denote the remainder of $k\in \mathbb{Z}$ modulo $661$, i.e., the unique integer $0 \le [k] < 661$ such that $[k] \equiv k \pmod{661}$. Since $1 \le b_n < 661$, it follows that $b_n = [b_n]$ for all $n\in \mathbb{N}$.
Lemma 1: Let $m \in \mathbb{Z}$ be an even integer. Then $[m/2] = [331m]$.
Proof: Note that $2$ is a unit in $\mathbb{Z}/661\mathbb{Z}$. Indeed, we have $2^{-1} \equiv 331 \pmod{661}$. Therefore we have
$$331m \equiv 331 \cdot \left[2\cdot \left(\frac{m}{2}\right)\right] \equiv [331 \cdot 2]\left(\frac{m}{2}\right)\equiv \frac{m}{2} \pmod{661}.$$
It follows that $[m/2] = [331m]$. $\square$
Lemma 2: For all $n\ge 1$, we have $b_n = [331^{(n-1)}]$.
Proof: Consider the defining recursion
$$b_{n+1} = \begin{cases}b_n/2 & 2 \mid b_n,\\ (b_n + 661)/2 & 2\not\mid b_n.\end{cases}$$
In the first case, we have
$$b_{n+1} = [b_{n+1}] = [b_n/2] = [331b_n].$$
In the second case, we have
$$b_{n+1} = [b_{n+1}] = [(b_n + 661)/2] = [331(b_n + 661)] = [331b_n].$$
In either case, we have $b_{n+1} = [331b_n]$. Starting with $b_1 = 1$, it follows that $b_n = [331^{(n-1)}]$. $\square$
The period of the sequence is therefore the order of $331$ mod $661$. The result then follows by noting $661$ is prime, so that $(\mathbb{Z}/661\mathbb{Z})^{\times} \cong \mathbb{Z}_{660}$ is cyclic, and moreover that $331$ (or equivalently, $2$) is a primitive root modulo $661$. This last fact can be verified with a quick (albeit tedious) calculation.
Best Answer
For part 2:
Hint: The question is essentially asking for the smallest positive integer $n$ such that $2^n \equiv 1 \pmod {125}$.
We know from corrected part 1 that $ 2^ {10} \equiv -1 \pmod{25}$. Now verify that this is the smallest $n$ for mod 25.
Hint: Hence conclude that $2^{100} \equiv 1 \pmod{125}$ is the smallest $n$.
So, the period is 100, starting from $a_3$.
For part 1, Brian's observation works directly.
Alternatively, you can easily modify the above.
(Sorry, my previous version thought you were still summing the last 3 digits in part 2)