Solutions

a)

Let $N\leq 0$. Then, for any $\delta>0$, we have

$$0<|x-3|<\delta \Rightarrow f(x)=\frac{1}{(x-3)^2}>0\geq N.$$

Now, let $N>0$. Pick

$$\delta=\frac{1}{\sqrt{N}}.$$

This gives

$$0<|x-3|<\frac{1}{\sqrt{N}} \Rightarrow \sqrt{N}<\frac{1}{|x-3|} \Rightarrow N<\frac{1}{(x-3)^2}.$$

b)

Suppose, by contradiction, that there is no $x$ such that $g(x)=0$. Suppose $f(x)>\varepsilon$ for all $x$ for some $\varepsilon>0$ and

$$\lim_{x\to a}g(x)=0.$$

Pick any $N$. Pick any $\alpha>0$ such that

$$\frac{\varepsilon}{\alpha}\geq N.$$

By definition, there exists $\delta>0$ such that

$$0<|x-a|<\delta \Rightarrow |g(x)|<\alpha.$$

Then

$$\frac{1}{\alpha}<\frac{1}{|g(x)|}$$

and therefore

$$N\leq \frac{\varepsilon}{\alpha}<\frac{\varepsilon}{|g(x)|}<\frac{f(x)}{|g(x)|}.$$