Is this a example of of two events occurring simultaneously?
No. It's "one by one" not "together/simultaneously". So, one ball is drawn first then depending upon the question it is replaced or not replaced into the box and then the next ball is drawn.
We can have $2$ balls drawn such that both are red. Better/correct word to use in this case would be "consecutively".
Now follow the below reasoning:
What if we were told that they were drawn simultaneously?
Let's say we are given that 2 balls are drawn simultaneously. What is the probability that both the balls are red?
If I'm picking $2$ balls simultaneously, then the $2$ balls can be simultaneously red too. We would solve it this way:
$M1:$ Long way
$E=\{(R_1R_2),(R_1R_3),(R_1R_4),(R_1R_5),(R_2R_3),(R_2R_4),(R_2R_5),(R_3R_4),(R_3R_5),(R_4R_5)\}$
$S=\{(W_1R_1),(W_1R_2),\dots,(R_3R_5),(R_4R_5)\}$
$n(E)=10, n(S)=36$
$P(E)=\frac{n(E)}{n(S)}=\frac{10}{36}=\frac5{18}$
$M2:$ Shorter way
$n(E)=$ $^5C_2=10$
$n(S)=$ $^9C_2=36$
$P(E)=\frac{n(E)}{n(S)}=\frac{10}{36}=\frac5{18}$
Note 1: "Simultaneously" means happening at same time. So there's no way one of the 2 balls, that is, the $1^{st}$ ball can be ever replaced. So, it would automatically mean that the replacement can't happen. That's why the $^nC_2$ worked.
Hope you can now work out yourself cases such as: $3$ dice are thrown smly, $4$ cards are picked smly from a standard deck, etc. where smly means simultaneously.
In some cases, it doesn't matter whether the experiment is one by one or simultaneously. Like that of rolling of multiple dice or tossing of multiple coins, etc. here the sample space will the same. (Basically those cases where replacing anyways can't happen).