Calculating probability of at least one occurrence in N trials when the probability of occurence is non-constant I've been asked to calculate the probability of an event occurring at least once in 50 trials (but let's leave it as N, just to be general), but with a twist I haven't seen before.
In the form :
For N attempts, the probability of event A occurring (that is, P(A)) on the first two attempts is 9/80 (and P(~A) = 71/80, just to clarify the probability space). On all subsequent attempts, P(A) = 13/2500 (P(~A) = 2487/2500). What is the probability that at least one trial observes event A?
I've done this kind of question many times in the past, but the probability of the event was always constant over all
trials. I'm not quite sure where to begin when the probability of the event occurring on trial N is a function of the number of trials elapsed.
 A: You don't get "two attempts" at an event $A$ in a probability space.
The event either occurs or does not.
What you can have is two or more events $A_1, A_2, \ldots,$
each of which may or may not occur.
For example, rolling a die three times, you can let $A_1$ be the event that you roll $1$ on the first roll,
$A_2$ be the event that you roll $1$ on the second roll,
$A_3$ be the event that you roll an odd number on the third roll.
Oops! Did you see what I did there? One of the three events has a different chance of occurring than the others.
Does that make $A_1, A_2,$ and $A_3,$ defined in this way,
independent or not independent?

Of course independence does matter for your question, but you have a more fundamental point of confusion so let's see if we can deal with that.
You are getting everything confused and mixed up due to using non-mathematical language in a mathematical setting and/or by using
mathematical formulas in a non-mathematical way.
In one place you write $P(A) = \frac{9}{80}.$ In another place you write $P(A) = \frac{13}{2500}.$ Either of those equations is fine on its own,
but when you try to use both equations in a single calculation, you are effectively saying that $\frac{9}{80} = \frac{13}{2500}.$
This is bound to lead to misunderstandings when other people try to understand what you are writing, and will probably also cause errors in your calculations.
Do not write like that.
Here's the setup. You are going to make $N$ attempts to see if a particular thing happens. It's too early to call this "event $A$" because we have not yet figured out what an "event" is.
So for the sake of exposition, I'll suppose you are throwing a dart at a wall and the thing you are interested in is whether the dart hits a particular dartboard hanging on the wall.
So $N$ attempts means throw the dart $N$ times and see if you hit the dartboard.
(For simplicity, I'll also assume that you keep throwing until you have tried $N$ times, even if you hit on one of the earlier throws. We could work things out assuming you stop after the first hit, but it gets more complicated to keep track of, and it does not affect the probability of whether you get at least one hit.)
Now, you might hit the dartboard on the very first attempt.
This is an event. If you hit the dartboard on the first attempt, then you hit it on the first attempt regardless of whether you hit it on any other attempt; conversely, if you miss on the first attempt, you missed on the first attempt regardless of what happened at any other time.
Let's give a name to this event: call it $A_1.$ That is, $A_1$ occurs if and only if you hit the dartboard on your first attempt.
Now you might or might not hit the dartboard on the second attempt.
If you hit on the second attempt, you hit on the second attempt
regardless of what might happen on any of the other attempts,
including the first attempt.
Let's give the name $A_2$ to the event that occurs if and only if you hit the dartboard on the second attempt.
Similarly for the third attempt, there is an event $A_3$ defined similarly to $A_1$ and $A_2,$ except $A_3$ depends only on whether you hit the dartboard on the third attempt, regardless of the first attempt, second attempt, or any other attempt.
Continue onward in this fashion, defining $A_4,$ $A_5,$ and so forth,
until finally you have defined an event that happens if and only if you hit the  dartboard on the $N$th attempt, and you have named that event $A_N.$
Now it is also possible to talk about the event, "I hit the dartboard at least once during the $N$ attempts, and it does not matter which attempt." You can even decide that the name of this event is $A.$
In the language of events in probability,
$$
A = A_1 \cup A_2 \cup A_3 \cup \ldots \cup A_N.
$$
(If you would rather not use the symbol $A$ for this, fine, call it something else. I used $A$ because there is nothing else in this answer for which I need to use the symbol $A$ without a subscript.)
According to the way you have tried to describe the problem, it appears that the probability of hitting the dartboard on the first attempt is
$$P(A_1) = \frac{9}{80}.$$
Likewise the probably of hitting on the second attempt is
$$P(A_2) = \frac{9}{80}.$$
But the probability of hitting on the third attempt is
$$P(A_3) = \frac{13}{2500}.$$
Likewise for all the later attempts,
$$P(A_4) = P(A_5) = \cdots = P(A_N) = \frac{13}{2500}.$$
Now it becomes important whether the $N$ events based on the $N$ individual throws are independent. If they are independent you can use a simple formula for the conjunction of independent events, which implies that
$$
P(\lnot A_1 \cap \lnot A_2 \cap \cdots \cap \lnot A_N)
= P(\lnot A_1) P(\lnot A_2) \cdots P(\lnot A_N).
$$
Then use De Morgan's Law,
$$
\lnot A_1 \cap \lnot A_2 \cap \cdots \cap \lnot A_N = 
\lnot(A_1 \cup A_2 \cup \cdots \cup A_N)
$$
and the fact
$$
P(\lnot B) = 1 - P(B).
$$
