If $ab=ba$ what are the possible orders of $ab$? The question is:
Let $a$ and $b$ be two elements of in a finite group $G$, say $o(a) = m$ and $o(b) = n$. If $ab = ba$, determine all possible values of $o(ab)$.(Assume $o(x)=$the order of x)
Proof. Assume $(m,n)=d$, $m=dm'$ and $n=dn'$, in the way that $(m',n')=1$.
$$\begin{align}
&d = p_1^{\alpha_1}\dots p_a^{\alpha_a}q_1^{\beta_1}\dots q_b^{\beta_b}t_1^{\gamma_1} \dots t_c^{\gamma_c}\\
&n' = p_1^{\alpha_1'}\dots p_a^{\alpha_a'}.n'', (n'',d)=1, \alpha_i' \ge 1\\
&m' = q_1^{\beta_1'}\dots q_b^{\beta_b'}.m'', (m'',d)=1, \beta_i' \ge 1
\end{align}$$
In fact, we break $d$ to three parts. First, primes that $n'$ posses. Second, primes that $m'$ posses. We use the fact that these two numbers do not have any primes in common too.($(n',m')=1$). 
We want to find for what $r$ we have: $(ab)^r=1$
$$\begin{equation}
ab=ba \rightarrow (ab)^r=a^rb^r=1\implies a^r=(b^{-1})^r
\end{equation}\tag{1}\label{eq1}
$$
We know $b$ has order n then $b^{-1}$ has the same order too. If we have $a^r=(b^{-1})^r$, so the we should have $o(a^r)=o((b^{-1})^r)$. Moreover, We know order of elements in the cyclic group $\langle a \rangle$ and $\langle b^{-1} \rangle$.
$$\begin{equation}
\begin{cases}
o(a^r)=\frac{m}{(m,r)}\\[2ex]
o((b^{-1})^{r})=\frac{n}{(n,r)}
\end{cases}
\xrightarrow{\eqref{eq1}} \frac{m}{(m,r)}=\frac{n}{(n,r)}\implies m.(n,r)=n.(m,r)
\end{equation}\tag{2}\label{eq2}
$$
$$\begin{equation}
m.(n,r)=n.(m,r) \rightarrow dm'.(dn',r)=dn'(dm',r)\implies m'.(dn',r)=n'(dm',r)
\end{equation}\tag{3}\label{eq3}
$$
As a result, we should have $m' |(dm',r)$, and this means that $m' |r$. With the symmetry:
$$\begin{equation}
\begin{cases}
m' |r\\[2ex]
n' |r
\end{cases}
\xrightarrow{(m',n')=1} n'm'|r \implies r=n'm'r'
\end{equation}\tag{4}\label{eq4}
$$
$$\begin{equation}
m'.(dn',r)=n'(dm',r) \xrightarrow{ r=n'm'r'} (d,r'm')=(d,r'n')
\end{equation}\tag{5}\label{eq5}
$$
Lemma 1:$ r|dn'm'$.
Proof. As we have $(ab)^{dn'm'}=a^{nm'}b^{mn'}=1_G$. Thus $o(ab)|dn'm'$.$\blacksquare$
As we proved that $r=n'm'r'$, and by the use of Lemma 1:
$$\begin{equation}
r=n'm'r'|dn'm' \rightarrow r'|d
\end{equation}\tag{6}\label{eq6}
$$
Consequently, we can write $r'$ in the following format:
$$\begin{equation}
r'=p_1^{\zeta_1}\dots p_a^{\zeta_a}q_1^{\eta_1}\dots q_b^{\eta_b}t_1^{\mu_1} \dots t_c^{\mu_c}\\
 \zeta_i \le \alpha_i , \eta_i \le \beta_i,\mu_i \le \gamma_i \\
\end{equation}\tag{7}\label{eq7}
$$
Now we want to use the $\eqref{eq5}$ to prove that $\zeta_i=\alpha_i$ and $\eta_i=\beta_i$.
$$\begin{equation}
\begin{cases}
 (d,r'm')= p_1^{\min(\alpha_1,\zeta_1)} \dots p_1^{\min(\alpha_a,\zeta_a)} q_1^{\min(\beta_1,\eta_1+\beta_1^{'})} \dots q_b^{\min(\beta_b,\eta_b+\beta_b^{'})} t_1^{\mu_1} \dots t_c^{\mu_c}\\[2ex]
 (d,r'n')= p_1^{\min(\alpha_1,\zeta_1+\alpha_1^{'})} \dots p_1^{\min(\alpha_a,\zeta_a+\alpha_a^{'})} q_1^{\min(\beta_1,\eta_1)} \dots q_b^{\min(\beta_b,\eta_b)} t_1^{\mu_1} \dots t_c^{\mu_c}\\
\end{cases}
\xrightarrow{\eqref{eq5}}
\end{equation}\tag{8}\label{eq8}
$$
$$\begin{equation}
\begin{cases}
\min(\alpha_i,\zeta_i+\alpha_i')=\min(\alpha_i,\zeta_i)= \zeta_i; \forall  1 \le i \le a\\[2ex]
\min(\beta_j,\eta_j+\beta_j')=\min(\beta_j,\eta_j) = \eta_j\; \forall  1 \le j \le b 
\end{cases} 
\xrightarrow{\beta_j',\alpha_i' \ge 1}
\begin{cases}
\alpha_i=\zeta_i\\[2ex]
\beta_j =\eta_j
\end{cases} 
\end{equation}\tag{9}\label{eq9}
$$
In the above equation we use the facts that $r'|d$ and $(m'',d)=(n'',d)=1$.
At the end we can conclude the following statement:
If $ab=ba$, $o(a)=m$ and $o(b)=n$. If $o(ab)=r$ then $r$ is the following format:
$$\begin{align}
&d = p_1^{\alpha_1}\dots p_a^{\alpha_a}q_1^{\beta_1}\dots q_b^{\beta_b}t_1^{\gamma_1} \dots t_c^{\gamma_c}\\
&n' = p_1^{\alpha_1'}\dots p_a^{\alpha_a'}.n'', (n'',d)=1, \alpha_i' \ge 1\\
&m' = q_1^{\beta_1'}\dots q_b^{\beta_b'}.m'', (m'',d)=1, \beta_i' \ge 1\\
&\implies r=n'm'p_1^{\alpha_1}\dots p_a^{\alpha_a}q_1^{\beta_1}\dots q_b^{\beta_b}t_1^{\mu_1} \dots t_c^{\mu_c}, \mu_i \le \gamma_i
\end{align}$$
Questions:

*

*Is this correct or am I missing something?

*Are any other restrictions that can be added?

*Is there any way to check this orders exist or not?

 A: I'm having a hard time following your argument/notation. It is simpler if we work one prime at a time.
Say $m=p^{\rho}$, $n=p^{\sigma}$, and without loss of generality assume that $0\leq\rho\leq \sigma$.
It is clear that the order of $ab$ divides $p^{\sigma}$, since $a^{p^{\sigma}}=b^{p^{\sigma}}=1$.
Claim 1. If $\rho\lt\sigma$, then the order of $ab$ is $\mathrm{lcm}(m,n) = p^{\sigma}$.
Proof. Suppose that $(ab)^{p^t} = e$, with $t\leq \sigma$. Then $b^{p^t}=a^{-p^t}$. Now, the order of $a^{p^t}$ is $\max(p^{\rho-t},1)$, while the order of $b^{p^t}$ is $p^{\sigma-t}$. Since $\rho-t\lt\sigma-t$ and $1\leq p^{\sigma-t}$, the only way we can have $\max(p^{\rho-t},1)=p^{\sigma-t}$ is if $1=p^{\sigma-t}$, which means $t=\sigma$, as required. $\Box$
You can even achieve this within the same cyclic group: take $C_{p^{\sigma}}$, cyclic group of order $p^{\sigma}$ generated by an element $z$; take $b=z$, and $a=z^{p^{\sigma-\rho}}$.
Claim 2. If $\rho=\sigma$, then the order of $ab$ may be any of $p^t$, $0\leq t\leq \sigma$.
Proof. Fix $t$ with $0\leq t\leq \sigma$. Let $C_{p^t}$ be the cyclic group of order $p^t$ with generator $x$, and let $C_{p^{\sigma}}$ be the cyclic group of order $p^{\sigma}$ with generator $y$. Let $a=(x,y)$ and $b=(1,y^{-1})$. Then both $a$ and $b$ have order $p^{\sigma}$, and $ab=(x,1)$ has order $p^t$, as required. $\Box$

Now, if I follow your notation correctly: the $p_i$ are the primes that occur in both $n$ and $m$, but occur in $m$ to a strictly smaller power than they do in $n$. The $q_i$ are the primes that occur in both $n$ and $m$, but occur in $n$ to a strictly smaller power than they do in $m$. And the $t_i$ are the primes that occur in both, to the exact same power. Meanwhile $n''$ consists of primes that occur in $n$ but not in $m$ (so, to a "different power"), and symmetrically with $m''$.
So, doing it one prime at a time as I do above, you get that your only leeway is in the power of the $t_i$, and those may be any quantity between $0$ and the largest power that divides $\gcd(n,m)$.
This is exactly what you have at the end. I can't quite follow all your calculations (too many indices, subindices, etc, too little time), but the answer to 2 is "No, that's the only restrictions you get", and the answer to 3 is "You construct explicit examples."
I've shown above how to construct specific examples for each prime, so you can then just take the direct product of each of the examples to get an example for orders $n$ and $m$.
