a) From $A^3 = B^3$ we get $\det A = \det B = d$ and by Hamilton-Cayley theorem we obtain $A^2 = aA - dI_2$, $B^2 = bB - dI_2$, where $a = \operatorname{tr} A$, $b = \operatorname{tr} B$. As a consequence one can write $A^3 = aA^2 - dA = (a^2 - d)A - adI_2$ and $B^3 = (b^2 - d)B - bdI_2$, so $(a^2 - d)A - adI_2 = (b^2 - d)B - bdI_2$ (*).

By right and left multiplication with $B$ we get $(a^2-d)BA - a dB = (b^2-d)B^2 - bdB = (a^2-d)AB - adB$, so $d = a^2$. In the same way, $d = b^2$, and by (*), we deduce $d = 0$ or $a = b$.
If $d=0$, then $a=b=0$, so $A^2 = B^2 = O_2$, from where $A^n = B^n$, for any $n \ge 2$. So $\text{tr } A = a = 0 = b = \text{tr } B$ and $\text{tr } A^n = \text{tr } B^n$, for all $n \ge 2$.
If $d \neq 0$, then $a = b$ and $C^2 = aC - a^2I_2$, $C^3 = -a^3I_2$, so $C^{3k} = (-a^3)^k I_2$, $C^{3k-2} = (-a^3)^{k-1}C$, $C^{3k-1} = (-a^3)^{k-1}(aC - a^2I_2)$, $\forall C \in \{A, B\}, \forall k \ge 1$.

b) For $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$, we have $AB = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$, $BA = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$,
$A^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$, $B^n = \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix}$ and $\text{tr } A^n = \text{tr } B^n = 2, \forall n \ge 1$.