It is known that for a laboratory computing system the number of system failures during a month has a Poisson distribution with a mean of 0.8. The system has just failed. Find the probability that at least 2 months will elapse before a further failure.
So I thought that I should find the $P(X=0)+P(X=1)$ and I found .898658 am I right?
I'm not sure why you're adding $P(X=0)$ and $P(X=1)$. Because there are two months?
The right approach is to consider what must happen in those two months. In the first month, there must be no failures. This happens with probability $P(X=0) = e^{-0.8}$. That must be repeated for a second month with no failures, which has the same probability. Since the two months are independent (a property of the Poisson distribution), the probability that at least two months will pass before the next failure is $[P(X=0)]^2 = e^{-1.6} \doteq 0.20190$.
One can also arrive at the same answer by considering that the exponential time to failure associated with a Poisson distribution with rate $\lambda = 0.8$ has a CDF of $F(t) = 1-e^{-0.8t}$. This represents the probability that the next failure happens by time $t$ months. We want the probability that the next failure does not happen within $t=2$ months, and this probability is $P(t>2) = 1-P(t<2) = 1-F(2) = e^{-1.6}$, as before.