My question is, would $r_1 =r_2$?
That's usually up to you as the designer of the system.
Usually when you declare that you have "a deterministic environment", you imply that both $s'$ and $r$ are fixed values depending on $(s,a)$. So in your examples, you would expect your observations to also have $r_1 = r_2$
However, it is possible to define a MDP where transition to state $s'$ is deterministic, but $r$ is not. For instance, you could define reward in a game equal to the sum of a number of dice rolled, with better rewards (on average) resulting in more dice. This is still a valid MDP and can be solved using RL techniques.
A real-world example of this might be managing a queue of work, where you want to minimise lead time, but don't know for certain how long each task will take. Your state progression moves deterministically - you have a queue of pending tasks, current tasks and workers, and assigning a task to a worker is completely deterministic. However, you don't know how efficiently tasks will be performed until after they are done, so you don't know the reward perfectly from the assignment (whether you can treat this as random or hidden state is a more complex issue - it is often pragmatic to treat such unknown data as random though).