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Solution:

Part a

Let $\epsilon>0$. Since $\lim a_n=s$, there exists $N_1>0$ such that
$$
|a_n-s|<\epsilon\Longrightarrow -\epsilon < a_n-s
$$ for all $n>N_1$. Similarly, since $\lim b_n=s$, there exists $N_2>0$ such that
$$
|b_n-s|<\epsilon\Longrightarrow b_n -s< \epsilon
$$ for all $n>N_2$.

Let $N=\max\{N_1,N_2\}$. Then we have
$$
s_n-s\le b_n-s<\epsilon
$$ and
$$
-\epsilon < a_n-s\le s_n-s
$$ for all $n>N$. Therefore, we have $|s_n-s|<\epsilon$ for all $n>N$. Hence $\lim s_n=s$ as desired.

Part b

Let $\epsilon>0$. Since $\lim t_n=0$, there exists $N>0$ such that
$$
|t_n-0|<\epsilon\Longrightarrow t_n<\epsilon
$$ for all $n>N$. Therefore, $n>N$ implies that
$$
|s_n-0|=|s_n|\le t_n<\epsilon
$$ as desired. Thus $\lim s_n=0$.