(a) $a_{n+1} =\frac{a_n +2}{a_n + 1}$ $\frac43 \le \frac{a_1 +2}{a_1 + 1} \le \frac32$ $4a_1+4 \le 3a_1+6 \implies a_1\le 2$ and $2a_1+4 \le 3a_1+3 \implies 1\le a_1$, thus $2\geq a_1\geq 1$. Suppose $\frac43 \le a_{n}\le \frac32$, then we want to show that for all $n\geq 1$ we have $\frac43 \le 1+\frac{1}{a_n + 1} \leq \frac{3}{2} \Longleftrightarrow 1 \leq a_n\leq 2$, which is true. Thus, $\boxed{a_1=[1,2]}$. (b) $a_{n+1} =\frac{a_n -3}{a_n + 1}$ We want to show that $a_4=a_1$, then $a_4=\frac{\frac{a_2 -3}{a_2 + 1}-3}{\frac{a_2 -3}{a_2 + 1} + 1}=\frac{\frac{a_2 -3-3(a_2+1)}{a_2 + 1}}{\frac{a_2 -3+(a_2+1)}{a_2 + 1}}=\frac{a_2 -3-3(a_2+1)}{a_2 -3+(a_2+1)}=\frac{-2a_2-6}{2a_2 -2}=-\frac{a_2+3}{a_2 -1}=-\frac{\frac{a_1 -3}{a_1 + 1}+3}{\frac{a_1 -3}{a_1 + 1}-1}=-\frac{\frac{a_1 -3+3(a_1+1)}{a_1 + 1}}{\frac{a_1 -3-1(a_1+1)}{a_1 + 1}}=-\frac{a_1 -3+3(a_1+1)}{a_1 -3-1(a_1+1)}=-\frac{4a_1}{-4}=a_1$. Notice that we cannot have $a_2=-1\Longleftrightarrow a_1=1$ or $a_1=-1$. Hence, we have showed that $a_4=a_1$, hence $a_{3k+1}=a_1$ and we are done.