Currently, whenever I encounter Poisson questions in the form of $P(X\ge x)$, I am forced to use
$P(X\ge x) =1-P(X\lt x)$
$1 - (P(X= 0) + P(X= 1) + P(X= 2) + ...+ P(X= x-1))$.
If in the future, I get a question where the given $x$ is large. Then I can't sum them individually, but I don't know how to solve the infinite sum or maybe there is other methods?
i.e.
Poisson(2)
$P(K\ge100)$
$=\sum_{k=100}^{\infty}\frac{e^{-2}2^k}{k!}$
$=$ ... ?
An example of normal appoximation:
Let $X \sim \mathsf{Pois}(\lambda = 100);$ find $P(X > 90):$ $$P(X > 90) = 1-P(X \le 90) = 1-P(X \le 90.5) = 1-P\left(\frac{X-\lambda}{\sqrt{\lambda}} \le \frac{90.5 - 100}{10}\right)\\ \approx 1-P(Z \le -0.95) = 0.8289,$$
where $Z$ is standard normal. The computation in R statistical software below shows the exact answer:
Nowadays, it seems that software is the practical way to compute many Poisson probabilities. In my view the normal approximation, relying as it does on printed tables or software, is not more genuine mathematics than using readily available software. However, for proving some theorems, relationships among normal, gamma, binomial, and Poisson distributions can be very useful.