This is a rather long answer, so feel free to skip over to the formula.
$\def\ord{\text{ord}}\def\div{\text{ divides }}\def\v#1#2{v_{#1}(#2)}$
Let $G$ be an abelian group with two elements $a,b$.
Suppose $a$ has order $m$ and $b$ has order $n$.
Denote $l=\text{lcm}(m,n)$ and $g=\text{gcd}(m,n)$.
Also define $H=\langle a\rangle,K=\langle b\rangle$ and $s=|H\cap K|$.
Our goal is to determine the order of $ab$.
Since $H\cap K$ is a subgroup of $H$ and $H$ is cyclic, we know
$H\cap K$ must be generated by $a^{m/s}$. Similarly $H\cap K$ is
generated by $b^{n/s}$. Hence there is some $k_0$ which is relatively prime to $s$ such that $a^{m/s}=b^{nk_0/s}$.
Lemma.
$$\frac ls\div\ord(ab)\div l.$$
Proof.
Let $x=\ord(ab)$. Since $(ab)^x=e$, we see
$a^x\in H\cap K=\langle a^{m/s}\rangle$, so $m/s\mid x$, i.e. $m\mid xs$.
Similarly, $n\mid xs$. Hence $l\mid xs$, namely, $l/s\mid\ord(ab)$.
Also, $(ab)^l=e$, so $\ord(ab)\mid l$.
Theorem.
$$\ord(ab)=\frac l{\gcd(s,\frac{m+nk_0}g)}.$$
Proof.
From the above lemma we know $\ord(ab)=lk/s$ for some $k$.
Now we consider $(ab)^{lk/s}=e$.
$$
\eqalign{
(ab)^{lk/s}&=(a^{m/s})^{lk/m}b^{lk/s}\cr
&=(b^{nk_0/s})^{lk/m}b^{lk/s}\cr
&=b^{lk/s(1+nk_0/m)}\cr
&=b^{lk(m+nk_0)/ms}
}
$$
So $(ab)^{lk/s}=e$ is equivalent with $n\mid lk(m+nk_0)/ms$, i.e. $mns\mid lk(m+nk_0)$.
Note that $mns=lgs$, so it is further equivalent with $s$ dividing $k(m+nk_0)/g$.
Let $g_1=\gcd(s,\frac{m+nk_0}g)$. Then it is equivalent with $s/g_1$ dividing $k\frac{m+nk_0}{g}/g_1$.
But we know $s/g_1$ and $\frac{m+nk_0}{g}/g_1$ are relatively prime, so this is equivalent with
$s/g_1$ dividing $k$, that is to say, the least such $k$ is $s/g_1$.
Therefore $\ord(ab)=lk/s=\frac{l(s/g_1)}s=\frac l{g_1}$ as desired.
From this formula we can derive a necessary and sufficient condition for $\ord(ab)=l$: $\gcd(s,\frac{m+nk_0}g)=1$. A non-trivial special case is when, for every prime divisor $p$ of $s$, $p$ does not divide $\frac{m+nk_0}g$, e.g. when $p$ divides $m$ and $n$ to different degrees.
Hope this helps.
Hint For $b)$ your approoach is not working since the converse to FLT is not true.
Try instead the following:
$\text{ord}_n (a)=n-1$ implies that $a, a^2,... , a^{n-1}$ are distinct elements modulo n, in the set $\{1, 2, .., n-1\} \pmod{n}$.
Deduce that $1, 2,.., n-1$ are all invertible modulo n
Best Answer
Note that Zsigmondy's theorem could be used to prove this. Instead, we can continue with your idea of using the multiplicative order. For each prime $r$ where $r \mid a^{pq}-1$, and their multiplicative orders $m = \operatorname{ord}_r(a)$, we have
$$m \mid pq \;\;\to\;\; m \in \{1, p, q, pq\}$$
With $m = 1$, consider
$$a^{pq}-1=(a-1)(a^{pq-1}+a^{pq-2}+\ldots+a+1)$$
Since $r \mid a - 1 \;\to\; a \equiv 1 \pmod{r}$, then
$$a^{pq - 1} + a^{pq - 2} + \ldots + a + 1 \equiv 1 + 1 + \ldots + 1 + 1 \equiv pq \pmod{r}$$
but $r \neq p$ and $r \neq q$, so
$$r \nmid a^{pq - 1} + a^{pq - 2} + \ldots + a + 1$$
Thus, the product of all primes $r$ (including multiplicities), where $r \mid a^{pq} - 1$ with $m = 1$, divides $a - 1$ (note the product is actually $a - 1$). Next, with $m = p$, use
$$(a^{p})^q - 1 = (a^{p} - 1)\left((a^{p})^{q - 1} + (a^{p})^{q - 2} + \ldots + a^{p} + 1 \right)$$
Since $r \mid a^{p}-1 \;\to\; a^{p} \equiv 1 \pmod{r}$, then
$$(a^{p})^{q - 1} + (a^{p})^{q - 2} + \ldots + a^{p} + 1 \equiv 1 + 1 + \ldots + 1 + 1 \equiv q \pmod{r}$$
but $r \neq q$, so
$$r \nmid (a^{p})^{q - 1} + (a^{p})^{q - 2} + \ldots + a^{p} + 1$$
This means that the product of all primes $r$ (including multiplicities), with $m = p$, divides $a^{p} - 1$. Similarly, for all primes $r$ where $m = q$, we get
$$r \nmid (a^{q})^{p - 1} + (a^{q})^{p - 2} + \ldots + a^{p} + 1$$
Since $a - 1 \mid a^p - 1$, and the primes $r$ where $m = 1$ and those where $m = p$ are distinct (because $p \neq 1$), then the product of all primes $r$, including multiplicities, with $m = 1$ or $m = p$, divide $a^p - 1$. Also, the product of all primes $r$, including multiplicities, with $m = q$, divide $a^q - 1$. Thus, the product of all the primes $r$ (including multiple instances of them) where $r \mid a^{pq} - 1$ with $m \in \{1, p, q\}$ can at the most be $(a^p - 1)(a^q - 1)$. However, with $p$ and $q$ being distinct primes, we have $pq \gt p + q$ so, since $a \gt 1$, then
$$\begin{equation}\begin{aligned} (a^p - 1)(a^q - 1) &= a^{p + q} - a^p - a^q + 1 \\ & \lt a^{p+q} - 1 \\ & \lt a^{pq} - 1 \end{aligned}\end{equation}$$
Thus, there must be at least one prime $r$ where $m = pq$. With (as you've already indicated) $m \mid r - 1$, we therefore have
$$pq \mid r - 1$$