a. First, we determine for what starting values $a_1 > 0$ the inequalities $\frac{4}{3} \le a_2 \le \frac{3}{2}$ hold. Then, we will prove that for those starting values, the inequalities $\frac{4}{3} \le a_n \le \frac{3}{2}$ are also valid for all $n \ge 2$.

First, we observe that $a_2 = \frac{a_1+2}{a_1+1}$ and that the denominator, $a_1 + 1$, is positive (since $a_1 > 0$). The inequality
$$\frac{4}{3} \le a_2 = \frac{a_1 + 2}{a_1 + 1} \le \frac{3}{2},$$
is therefore equivalent to the inequality
$$\frac{4}{3}(a_1 + 1) \le a_1 + 2 \le \frac{3}{2}(a_1 + 1),$$
as we can multiply all parts in the inequality by the positive number $a_1 + 1$. Subtracting $a_1 + 2$ from all parts of the inequality, we see that this is equivalent to
$$\frac{1}{3}a_1 - \frac{2}{3} \le 0 \le \frac{1}{2}a_1 - \frac{1}{2}.$$
We therefore need to have $\frac{1}{3}a_1 \le \frac{2}{3}$ (i.e. $a_1 \le 2$), and $\frac{1}{2} \le \frac{1}{2}a_1$ (i.e. $1 \le a_1 \le 2$). The starting value $a_1$ must therefore satisfy $1 \le a_1 \le 2$.

Now suppose that $1 \le a_1 \le 2$, so that $a_2$ satisfies $\frac{4}{3} \le a_2 \le \frac{3}{2}$. Looking at $a_3$, we see that $a_3 = \frac{a_2+2}{a_2+1}$. That is the same expression as for $a_2$, only with $a_1$ replaced by $a_2$. Since $a_2$ also satisfies $1 \le a_2 \le 2$, the same argument now shows that $\frac{4}{3} \le a_3 \le \frac{3}{2}$.

We can repeat the same argument to show this for $a_4$, $a_5$, etcetera. Hence, we find that $\frac{4}{3} \le a_n \le \frac{3}{2}$ holds for all $n \ge 2$. The formal proof is done using induction: the induction basis $n = 2$ has been shown above. For the induction step, see the solution of part (b) of the version for klas 5 & klas 4 and below. The result is that all inequalities hold if and only if $1 \le a_1 \le 2$.

b. Let's start by computing the first few numbers of the sequence in terms of $a_1$. We see that
$$a_2 = \frac{a_1 - 3}{a_1 + 1}$$
and
$$\begin{aligned} a_3 &= \frac{a_2 - 3}{a_2 + 1} = \frac{\frac{a_1-3}{a_1+1} - 3}{\frac{a_1-3}{a_1+1} + 1} = \frac{a_1 - 3 - 3(a_1 + 1)}{a_1 - 3 + (a_1 + 1)} \\ &= \frac{-2a_1 - 6}{2a_1 - 2} = \frac{-a_1 - 3}{a_1 - 1}. \end{aligned}$$
Here, it is important that we do not divide by zero, that is, $a_1 \ne -1$ and $a_2 \ne -1$. The first inequality follows directly from the assumption. For the second inequality we consider when $a_2 = -1$ holds. This is the case if and only if $a_1 - 3 = -(a_1 + 1)$, if and only if $a_1 = 1$. Since we assumed that $a_1 \ne 1$, we see that $a_2 \ne -1$. The next number in the sequence is
$$a_4 = \frac{a_3 - 3}{a_3 + 1} = \frac{\frac{-a_1-3}{a_1-1} - 3}{\frac{-a_1-3}{a_1-1} + 1} = \frac{-a_1 - 3 - 3(a_1 - 1)}{-a_1 - 3 + (a_1 - 1)} = \frac{-4a_1}{-4} = a_1.$$
Again, we are not dividing by zero since $a_3 = -1$ only holds when $-a_1 - 3 = -a_1 + 1$, which is never the case.

We see that $a_4 = a_1$. Since $a_{n+1}$ only depends on $a_n$, we see that $a_5 = a_2$, $a_6 = a_3$, $a_7 = a_4$, etcetera. In other words: the sequence is periodic with period 3, and we see that
$$a_{2020} = a_{2017} = a_{2014} = \dots = a_4 = a_1.$$
To conclude: indeed we have $a_{2020} = a_1$ for all starting values $a_1$ unequal to 1 and $-1$.