For a given value $t$, we consider number sequences $a_1, a_2, a_3,...$ such that $a_{n+1} =\frac{a_n + t}{a_n + 1}$ for all $n \ge 1$.

(a) Suppose that $t = 2$. Determine all starting values $a_1 > 0$ such that $\frac43 \le a_n \le \frac32$ holds for all $n \ge 2$.

(b) Suppose that $t = -3$. Investigate whether $a_{2020} = a_1$ for all starting values $a_1$ different from $-1$ and $1$.
### Part (a)

Given the recurrence relation:
\[ a_{n+1} = \frac{a_n + 2}{a_n + 1} \]

We need to determine all starting values \( a_1 > 0 \) such that:
\[ \frac{4}{3} \le a_n \le \frac{3}{2} \]
holds for all \( n \ge 2 \).

1. **Initial Condition Analysis:**
   We start by analyzing the condition for \( a_2 \):
   \[ \frac{4}{3} \le \frac{a_1 + 2}{a_1 + 1} \le \frac{3}{2} \]

2. **Solving the Inequalities:**
   - For the left inequality:
     \[ \frac{4}{3} \le \frac{a_1 + 2}{a_1 + 1} \]
     \[ 4(a_1 + 1) \le 3(a_1 + 2) \]
     \[ 4a_1 + 4 \le 3a_1 + 6 \]
     \[ a_1 \le 2 \]

   - For the right inequality:
     \[ \frac{a_1 + 2}{a_1 + 1} \le \frac{3}{2} \]
     \[ 2(a_1 + 2) \le 3(a_1 + 1) \]
     \[ 2a_1 + 4 \le 3a_1 + 3 \]
     \[ 1 \le a_1 \]

3. **Combining the Results:**
   From the above inequalities, we get:
   \[ 1 \le a_1 \le 2 \]

4. **Inductive Step:**
   Suppose \( \frac{4}{3} \le a_n \le \frac{3}{2} \). We need to show that:
   \[ \frac{4}{3} \le a_{n+1} \le \frac{3}{2} \]
   Given:
   \[ a_{n+1} = \frac{a_n + 2}{a_n + 1} \]
   - If \( a_n = \frac{4}{3} \):
     \[ a_{n+1} = \frac{\frac{4}{3} + 2}{\frac{4}{3} + 1} = \frac{\frac{10}{3}}{\frac{7}{3}} = \frac{10}{7} \approx 1.428 \]
     which is between \( \frac{4}{3} \) and \( \frac{3}{2} \).

   - If \( a_n = \frac{3}{2} \):
     \[ a_{n+1} = \frac{\frac{3}{2} + 2}{\frac{3}{2} + 1} = \frac{\frac{7}{2}}{\frac{5}{2}} = \frac{7}{5} = 1.4 \]
     which is also between \( \frac{4}{3} \) and \( \frac{3}{2} \).

Thus, the sequence remains within the bounds for all \( n \ge 2 \).

Conclusion:
\[ \boxed{a_1 \in [1, 2]} \]

### Part (b)

Given the recurrence relation:
\[ a_{n+1} = \frac{a_n - 3}{a_n + 1} \]

We need to investigate whether \( a_{2020} = a_1 \) for all starting values \( a_1 \) different from \( -1 \) and \( 1 \).

1. **Analyzing the Recurrence Relation:**
   We need to show that \( a_4 = a_1 \).

2. **Calculating \( a_2 \), \( a_3 \), and \( a_4 \):**
   \[ a_2 = \frac{a_1 - 3}{a_1 + 1} \]
   \[ a_3 = \frac{a_2 - 3}{a_2 + 1} = \frac{\frac{a_1 - 3}{a_1 + 1} - 3}{\frac{a_1 - 3}{a_1 + 1} + 1} \]
   Simplifying \( a_3 \):
   \[ a_3 = \frac{\frac{a_1 - 3 - 3(a_1 + 1)}{a_1 + 1}}{\frac{a_1 - 3 + (a_1 + 1)}{a_1 + 1}} = \frac{\frac{a_1 - 3 - 3a_1 - 3}{a_1 + 1}}{\frac{a_1 - 3 + a_1 + 1}{a_1 + 1}} \]
   \[ a_3 = \frac{\frac{-2a_1 - 6}{a_1 + 1}}{\frac{2a_1 - 2}{a_1 + 1}} = \frac{-2a_1 - 6}{2a_1 - 2} = -\frac{a_1 + 3}{a_1 - 1} \]

   Now, calculate \( a_4 \):
   \[ a_4 = \frac{a_3 - 3}{a_3 + 1} = \frac{-\frac{a_1 + 3}{a_1 - 1} - 3}{-\frac{a_1 + 3}{a_1 - 1} + 1} \]
   Simplifying \( a_4 \):
   \[ a_4 = \frac{-\frac{a_1 + 3 + 3(a_1 - 1)}{a_1 - 1}}{-\frac{a_1 + 3 - (a_1 - 1)}{a_1 - 1}} = \frac{-\frac{a_1 + 3 + 3a_1 - 3}{a_1 - 1}}{-\frac{a_1 + 3 - a_1 + 1}{a_1 - 1}} \]
   \[ a_4 = \frac{-\frac{4a_1}{a_1 - 1}}{-\frac{4}{a_1 - 1}} = \frac{-4a_1}{-4} = a_1 \]

Thus, \( a_4 = a_1 \), and by induction, \( a_{4k+1} = a_1 \) for any integer \( k \).

Conclusion:
\[ \boxed{a_{2020} = a_1} \]