denominator with standard part 0 
(i) Why do we have to simplify the numerator and denominator, and not just substitute the standard part st(c) right away since it is given as 4.
(ii) Also what does this mean (more specifically, what is the idea behind the bold part):

However, since c $\neq$ 4 the fraction is defined, and it can be simplified by factoring the numerator and denominator

 A: (i) If you try to apply standard part rules directly you run into erroneous equations. For example, if you distribute the standard part map over the fraction you get
$$
st\big(\frac{c^{2}+2c-24}{c^{2}-16}\big)=\frac{st(c^{2}+2c-24)}{st(c^{2}-16)}\tag{1}
$$
Or maybe you try to do all possible distributions at once and get
$$
st\big(\frac{c^{2}+2c-24}{c^{2}-16}\big)=\frac{st(c)^{2}+2st(c)-24}{st(c)^{2}-16}\tag{2}
$$
But since $st(c)=4$, we get zero in the denominators of right side in both (1) and (2). So this doesn't make sense and the approach fails.
(ii) On the other hand, since $c^{2}\neq 16$, the nonstandard fraction
$$
\frac{c^{2}+2c-24}{c^{2}-16}\tag{3}
$$
is a well-defined nonstandard number. This what they mean by saying the fraction is defined. They go on to simplify $(3)$ to
$$
\frac{c+6}{c+4}\tag{4}
$$
Now the "direct approach" to evaluating the standard part (that we tried and failed to do in (i)) will work because the standard part of the denominator in $(4)$ is nonzero.
An analogy: You can liken this to certain limits in which one gets rid of $\frac{0}{0}$ by canceling common factors. For example consider
$$
\lim_{x\to 4} \frac{x^{2}+2x-24}{x^{2}-16}
$$
If you try to "plug in $4$" you get $\frac{0}{0}$. This is the analogy of the failed attempt in (i).
On the other hand we can factor and simplify:
$$
\lim_{x\to 4}\frac{x^{2}+2x-24}{x^{2}-16} = \lim_{x\to 4}\frac{(x+6)(x-4)}{(x+4)(x-4)} = \lim_{x\to 4}\frac{x+6}{x+4}=\frac{10}{8}.
$$
In the last step, plugging in $4$ causes no problems. This is the analogy of the successful approach in (ii).
