If ${\rm ord}(a) = n$ where $n$ is odd, then ${\rm ord}(a^2) = n$. $\DeclareMathOperator{\ord}{ord}$
Exercise 10.D.4 from Pinter says:

Let $a$ be any element of finite order of a group $G$. Prove the following:
If $\ord(a) = n$ where $n$ is odd, then $\ord(a^2) = n$.

Here's an approach I took.
We'll be using the following from the same chapter:
10.D.2

The order of $a^k$ is a divisor (factor) of the order of $a$.

10.T5 (Theorem 5)

Suppose an element $a$ in a group has order $n$. Then $a^t = e$ iff $t$ is a multiple of $n$.

Let's begin. By 10.D.2
$$\ord(a^2) \mid n$$
Let $m = \ord(a^2)$.
$$m \mid n$$
$$(a^2)^m = e$$
$$a^{2m} = e$$
By 10.T5
$$n \mid 2m$$
Due the the following rules:
$$even * even = even$$
$$even * odd  =  even$$
$$odd  * even = even$$
$$odd  * odd  = odd$$
$2m$ must be even.
But $n$ divides it so $n$ must be a factor of $m$.
Thus we have:
$$ m | n$$
$$  n | m$$
And so
$$m = n$$
Question 1: Is this an OK approach? I realize there are other approaches, but I wanted to explore one that only uses facts presented in or before this chapter in the book.
Question 2: This page also presents a similar solution.
He ends up with the following, like the above approach:
$$m | n$$
$$ n | 2m$$
At this point he states that:

$\gcd(n, 2) = 1$ (because n is odd)
Hence $n | m$

I understand the fact that $\gcd(n, 2) = 1$. But, how does he go from that fact to $n | m$?
 A: 
Question 1: Is this an OK approach?

The argument that $\, o(a^2) =: m\mid n\mid 2m\,$ is correct. But the inference $\,n\mid m\,$ is not properly justified. To  use parity, note $\, n\mid 2m\,\Rightarrow\, nk = 2m\,$ is even hence $k$ is even, by $\,n\,$ odd. So cancelling $2$ yields $\, n(k/2) = m,\,$ so $\, n\mid m\,$ as claimed.

At this point he states that:

gcd(n, 2) = 1 (because n is odd).  Hence n | m

I understand the fact that gcd(n, 2) = 1. But, how does he go from that fact to n | m?

Likely they apply Euclid's Lemma $\, \gcd(n,a)=1,\ n\mid am\,\Rightarrow\, n\mid m$ or it's generalization below.
Theorem $\, \ m\mid cx \iff\, \dfrac{m}{(m,c)}\ {\Large \mid}\ x.\ \ \,$ Proof $\,\ $ Let $\ d = (m,c).\ $  Then
we deduce $\, \ m\mid cx \overset{{\rm cancel}\ d\!\!}\iff\ \color{#c00}{\dfrac{m}d}\ {\Large \mid}\ \color{#c00}{\dfrac{c}d}\:x\!\!\overset{\rm(EL)\!}\iff\! \dfrac{m}d\ {\Large \mid}\ x\,\ $ by Euclid's Lemma (EL),
because: $\,\ (m,c) = d\ \Rightarrow\, \color{#c00}{\left(\dfrac{m}d,\,\dfrac{c}d\right)} = (m,c)/d = 1\ $ by the GCD Distributive Law
A: It is a generalization of Euclid's lemma that:  $a\mid bc \land (a,b)=1\implies a\mid c$.  (This is sometimes called Gauß's lemma.)
Or, use the fact from the theory of cyclic groups:   $\vert a^k\vert=\dfrac{\vert a\vert}{\operatorname {gcd}(\vert a\vert,k)}$.
A: Alternatively, write $n=2k+1$.
Then $a=ea=a^n a=a^{n+1}=a^{2k+2}=(a^2)^{k+1}$.
Therefore
$
\langle a^2\rangle
\subseteq
\langle a\rangle
\subseteq
\langle a^2\rangle
$
and so
$
\langle a^2\rangle
=
\langle a\rangle
$.
Thus,
${\rm ord}(a^2) = |\langle a^2\rangle| = |\langle a\rangle| = {\rm ord}(a)$.
