Suppose $a>0$.
I am given the following three equations: $$ \begin{align*} x_1'(t)&=e^{-ax_1(t)}-e^{-ax_2(t)}\\ x_2'(t)&=e^{-ax_2(t)}-e^{-ax_3(t)}\\ x_3'(t)&=e^{-ax_3(t)}, \end{align*} $$ where all three functions $x_i, i=1,2,3$, are non-negative.
The third equation can be solved explicitly (by separation of variables), $$ x_3(t)=\frac{1}{a}(\ln(a)+\ln(t+C)). $$ In particular, $x_3(t)\to\infty$ as $t\to\infty$.
Now I would like to prove that
(1) $x_1(t)\to\infty$ and $x_2(t)\to\infty$ as $t\to\infty$ ,
(2) there exists some $T>0$ such that $x_1(t)<x_2(t)<x_3(t)$ for all $t>T$.
Here is what I tried.
Proof of (1)
First show that $x_2(t)\to\infty$ as $t\to\infty$. Suppose by contradiction, that $x_2$ is bounded, i.e. there exists some $M>0$ such that $x_2\leq M$. Then, $x_2'(t)\geq e^{-aM}-e^{-ax_3(t)}$ and, for $t>0$ large enough, $x_2'(t)>0$ since $e^{-ax_3(t)}\to 0$. This implies that $x_2$ converges since $x_2(t)$ is bounded from above and, eventually, monotonically increasing. Consequently, $x_2'(t)\to 0$, implying $x_2(t)\to x_3(t)$. This contradicts the boundedness of $x_2$. Thus, $x_2(t)\to\infty$ for $t\to\infty$.
Next, since it is proven that $x_2(t)\to\infty$, the same argument holds for $x_1(t)$ which concludes the proof.
Proof of (2)
First show that there exists some $T>0$ such that $x_2(t)<x_3(t)$ for all $t>T$. To this end, assume by contradiction that for all $T>0$ there exists some $t'>T$ such that $x_2(t')\geq x_2(t')$, i.e. $x_2'(t')\leq 0$. Due to $x_2(t)\to\infty$ there also exists some $t''>t'$ such that $x_2'(t'')>0$. Since $x_3$ increases monotonically, this implies that $x_2$ oscillates around $x_3$. However, this is not possible since the nullcline $x_2=x_3$ can be crossed only in one direction. More precisely, the set $\{(x_2,x_3): x_2<x_3\}=\{(x_2,x_3): x_2'>0\}$ is forward invariant.
Hence, there exists some $T>0$ such that $x_2'(t)>0$ for all $t>T$.
Now, since $x_2$ increases monotonically for $t>T$, repeat the same argument to show that there exists some $T'>T$ such that $x_1(t)<x_2(t)$ for all $t>T'$. All together, statement (2) holds for $T'$.
I am very curious to know if my proofs are okay.
Both proofs looks alright, keep up with this! I should remark, that although both proofs are correct, I prefer the second one, because it’s more simple and easy to understand. One note, if you are practicing for a competition, try to be a little more specific in your answer.