Solution:

a) It follows from $a_{n}<1$ and $a_{n+1}=\frac{3}{a_{n}+2}<1$ that $a_{n}<-2$. Thus, $a_{n+1}<-2$, and therefore $a_{n}+2=\frac{3}{a_{n+1}}> -\frac{3}{2}$, i.e. $a_{n}>-\frac{7}{2}$.

б) First solution. Set $b_{n}=a_{n}+3$. Then $b_{n+1}=\frac{3 b_{n}}{b_{n}-1}$. It follows from a) that $-\frac{3}{2}<b_{n}-1<0$, i.e. $|b_{n}-1|<\frac{3}{2}$ and therefore $|b_{n+1}|>2|b_{n}|$. Hence $1>|b_{n+1}|>2^{n}|b_{1}|$ for all $n$. Letting $n \rightarrow \infty$ gives $b_{1}=0$. Then $b_{n}=0$, i.e. $a_{n}=-3$ for all $n$.

Second solution. Let $c_{1}=1$ and $c_{k+1}=3 c_{k}+(-1)^{k}$ for $k \geq 1$. One can prove by induction on $k$ that
$$-\frac{c_{2 k+1}}{c_{2 k}}<a_{n}< -\frac{c_{2 k}}{c_{2 k-1}}$$
for all $n$ and $k$. Indeed, similar to a) it follows from $a_{n}< -\frac{c_{2 k}}{c_{2 k-1}}$ that $-\frac{c_{2 k+1}}{c_{2 k}}< a_{n}$, which in turn implies that $a_{n}< -\frac{c_{2 k+2}}{c_{2 k+1}}$.
Further, induction on $k$ gives $c_{k}=\frac{3^{k}+(-1)^{k-1}}{4}$. Therefore $\lim_{k \rightarrow \infty} \frac{c_{k+1}}{c_{k}}=3$ and using (*) we conclude that $a_{n}=-3$ for all $n$.