Restaurant sending orders every 5 minutes on average, What's the probability in a particular day, that after the second order, an hour will pass until the next order sending.
I can conclude that's it's about Exponential distribution with $\lambda = 12$, because our time unit is hours, and we have 12 orders in one hour.
I'm not sure how to refer to the meaning of 'after the second order'.
What I've written so far is $Pr[X \ge 1] = 1 - Pr[X \le 1] = 1 - e^{-\lambda t}$ and $t = 1$
Your idea to compute $P(X > 1)$ where $W \sim \mathsf{Exp}(\lambda = 12)$ is OK, but you have made a simple mistake in writing the equation. Please look at it carefully. Your equation will give an answer $\approx 1.$
Intuitively, the probability of waiting more than an hour for the next order must be quite small--because you have already figured out that on average there are 12 orders an hour (one every five minutes).
You have correctly adjusted $\lambda$ and correctly applied the 'no-memory property' to look only one hour ahead (ignoring the past). So you pretty much understand what's going on. I'll let you find your simple mistake on your own. That way maybe you won't make it again on an exam or in an important application.
In R statistical software
pexpis an exponential CDF with the rate specified by the second argument. So the computation below gives the answer. However, this is just for reinforcement; you can easily compute the result on a simple calculator.Also, sometimes a simple sketch of the PDF helps avoid simple errors.