Using the fact that $(X_k)$ are identically distributed, we have
$$ \sum_{k=1}^{\infty} \Bbb{P}(A_k)
= \sum_{k=1}^{\infty} \Bbb{P}(X_1 > k \epsilon)
= \sum_{k=1}^{\infty} \Bbb{E}[\mathbf{1}_{\{ X_1 > k \epsilon \}}]
= \Bbb{E}\bigg[ \sum_{k=1}^{\infty} \mathbf{1}_{\{ X_1 > k \epsilon \}} \bigg]. \tag{*}$$
Now let us focus on the random variable $Y := \sum_{k=1}^{\infty} \mathbf{1}_{\{ X_1 > k \epsilon \}}$. It is not hard to see that
$$ Y = \begin{cases}
k, & X_1 \in (k\epsilon, (k+1)\epsilon] \text{ for some } k = 0, 1, \cdots \\
0, & X_1 = 0
\end{cases} $$
This proves that $Y \leq \frac{1}{\epsilon}X_1$ and hence
$$ \sum_{k=1}^{\infty} \Bbb{P}(A_k) = \Bbb{E}[Y] \leq \frac{1}{\epsilon}\Bbb{E}[X_1] < \infty. $$
Now all you have to do is to apply the 1st Borel-Cantelli's lemma.
Remark. If we replace the assumption $\Bbb{E}[X_1] < \infty$ by $\Bbb{E}[X_1] = \infty$, then a similar argument shows that $\limsup_{n\to\infty} \frac{X_n}{n} = \infty$ a.s. Indeed, one has the inequality
$$ Y \geq \frac{1}{\epsilon}X_1 - 1.$$
Plugging this back to the computation $\text{(*)}$, we have
$$ \sum_{k=1}^{\infty} \Bbb{P}(A_k) = \Bbb{E}[Y] \geq \frac{1}{\epsilon}\Bbb{E}[X_1] - 1 = \infty. $$
Therefore by the 2nd Borel-Cantelli lemma, $\Bbb{P}(\frac{X_n}{n} > \epsilon \text{ i.o.}) = 1$ and thus $\limsup_{n\to\infty} \frac{X_n}{n} \geq \epsilon$. Since $\epsilon > 0$ is arbitrary, the conclusion follows by letting $\epsilon \to \infty$ (along a subsequence, if you want to be rigorous).
The claim isn't true for $c<0$. For example, let $X$ be uniform (0,1), $c=-10$ and $t=1/c$.
For $c=0$, the claim is trivial. For $c>0$, we have:
$$
\Big[X>t\iff\exp(cX)>\exp(ct)\Big]\implies\Pr(X>t)=\Pr[\exp(cX)>\exp(ct)]
$$
Now use Markov's inequality with $Y=\exp(cX)$.
Best Answer
I think you need $m >1$ for this. Let $Y$ be a positive r.v. such that $P(Y \leq y)=1-e^{-2my^{2}}$. The density of $Y$ is given by $f(y)=4my e^{-2my^{2}}, y>0$. Since $P(X\geq y) \leq P(Y\geq y)$ we get $ P( Z \geq y) \leq P(e^{2(m-1)Y^{2}}\geq y)$. Hence, $EZ \leq Ee^{2(m-1)Y^{2}}$. An easy computetion shows that $Ee^{2(m-1)Y^{2}}=m$.