1).Prove: $\lim_{n \to \infty} \frac{3n^2-4}{n^2-4} = 3$. A proof: there exists an $\epsilon>0$. We define $N=\lceil{\sqrt{\frac{8}{\epsilon}+4}}\rceil$ (notice $N>2$). So: $$ \begin{aligned} |\frac{3n^2-4}{n^2-4}-3| < \epsilon \Leftrightarrow |\frac{3n^2-4-3n^2+12}{n^2-4}|<\epsilon \Leftrightarrow |\frac{8}{n^2-4}| < \epsilon \Leftrightarrow \frac{8}{n^2-4}<\epsilon \Leftrightarrow\\ 8<n^2\epsilon-4\epsilon \Leftrightarrow \frac{8+4\epsilon}{\epsilon}<n^2 \Leftrightarrow \sqrt{\frac{8}{\epsilon}+4}<n. \end{aligned} $$ This statement is true for all $n>N$, because: $$\lceil{\sqrt{\frac{8}{\epsilon}+4}}\rceil \geq \sqrt{\frac{8}{\epsilon}+4}.$$ thus, the limit is 3. $\Box$
2).$a_nb_n=1$. Prove or contradict: If $\lim_{n\to\infty}|a_n|=1$ so $\lim_{n\to\infty}|b_n|=1$. This statement is true. A proof: We choose $\epsilon = \frac{1}{2}$, so: $$ -\epsilon < a_nb_n-1 < \epsilon\Rightarrow -\epsilon+1<a_nb_n<\epsilon+1 \Rightarrow -\frac{1}{2}+1<a_nb_n<\frac{1}{2}+1 \Rightarrow \\ \frac{1}{2}<a_nb_n<1\frac{1}{2} $$ Also: $$ ||a_n|-1|<\epsilon \Rightarrow \frac{1}{2} <|a_n| <1\frac{1}{2} $$ It's obvious that $a_n\neq 0$ for almost all n, otherwise $\lim_{n\to\infty}a_n=0$. If $a_n>0$ for almost all n, also $b_n>0$ for almost all n. we can divide and get: $$ 1<\frac{a_nb_n}{|a_n|}<1 \Rightarrow1<\frac{a_nb_n}{a_n}<1 \Rightarrow 1<b_n<1 $$ If $a_n<0$ for almost all n, also $b_n<0$ for almost all n. we can divide and get: $$ 1<\frac{a_nb_n}{|a_n|}<1 \Rightarrow1<\frac{a_nb_n}{-a_n}<1 \Rightarrow 1<-b_n<1 $$ And from the sandwich theorem we get $\lim_{n\to\infty}|b_n| = 1$. $\Box$
As an alternative to simplify note that
$$\lim_{n \to \infty} \frac{3n^2-4}{n^2-4} = \lim_{n \to \infty} \frac{3n^2-12+8}{n^2-4} =\lim_{n \to \infty} 3+\frac{8}{n^2-4}$$
and we can show that eventually $0\le\frac1{n^2-4}\le\frac1n$ and $\frac1n\to 0$.