Let p be any prime, then
$\left(\frac{a}{p}\right) \equiv a^{(p-1)/2}$ (mod $p$)
For the case $p|a$, the textbook just says it is obviously.
If $p|a$, is $a^{(p-1)/2}\equiv0$ (mod $p$), am I correct?
From the definition, we know
Given $p$ is an odd prime, the Legendre symbol $\left(\frac{a}{p}\right)=0$ if $p|a$
But Why $\left(\frac{a}{p}\right) = 0$? In such case(i.e $(a, p) \neq 1)$, is $a$ neither a quadratic residue nor a quadratic nonresidue modulo m?
You are correct saying that if $p$ divide $a$, then $a^{(p-1)/2}\equiv 0\bmod p$. In fact, $a\equiv 0\bmod p$, which is stronger.
Regarding your other question, $a$ is either a quadratic residue $\bmod p$ or a quadratic nonresidue $\bmod p$. This is exactly the same saying that your jacket is either blue or not blue. The point is that $\displaystyle\left(\frac{a}{p}\right)=1$ if and only if $a$ is a nonzero quadratic residue $\bmod p$.
As I am not sure what you are exactly asking, let me know if I misunderstood something.