If $X_1, X_2,..$ is a sequence of independent random variables, each exponential such that $\mathbb{E}[X_n] = \frac{1}{\lambda_n}$.
1) If $\displaystyle \sum_{n=1}^\infty \frac{1}{\lambda_n} < \infty $ then $\mathbb{P}(\sum_{n=1}^\infty X_n < \infty ) = 1 $.
2) If $\displaystyle \sum_{n=1}^\infty \frac{1}{\lambda_n} = \infty $ then $\mathbb{P}(\sum_{n=1}^\infty X_n = \infty ) = 1 $.
This seems to be a Kolmogorov three series application. Kolmogorov Three Series Theorem
For 1)
Let $Y = \sum_{n=1}^\infty X_n$. The partial sums are be monotone increasing and each $X_n$ has support on $[0,\infty)$ and $X_n \ge 0$. We have $\sum_{n=1}^N X_n = Y_N \to Y$.
By monotone convergence theorem:
$\mathbb{E}(\sum_{n=1}^\infty X_n) = \sum_{n=1}^\infty \mathbb{E}(X_n)$.
Then applying Markovs inequality:
$\displaystyle \mathbb{P}(Y > n) \le \frac{\mathbb{E}(Y)}{n} \to 0$.
For 2)
I was thinking the Borel-Cantelli lemma but wasn't sure how to set it up.
Kolmogorov 3-series theorem is not needed here. If $\sum_{n=1}^\infty \frac1{\lambda_n}<\infty$, then $$\sum_{n=1}^\infty \mathbb E[X_n] = \mathbb E\left[\sum_{n=1}^\infty X_n\right]<\infty,$$ and because $\sum_{n=1}^\infty X_n\geqslant 0$, we have $\sum_{n=1}^\infty X_n<\infty$ with probability one.
This of course is not a complete answer, but was rather large for a comment.