Each of $3600$ subscribers of a telephone exchange calls it once an hour on average. What is the probability that in a given second $5$ or more calls are received? Estimate the mean interval of time between such seconds $(i, i + 1)$.
1. **Modeling the Problem:**
   - We are given that each of the 3600 subscribers calls once an hour on average.
   - This translates to a total of \(3600\) calls per hour.
   - Since there are \(3600\) seconds in an hour, the average rate of calls per second is:
     \[
     \lambda = \frac{3600 \text{ calls}}{3600 \text{ seconds}} = 1 \text{ call/second}
     \]
   - We model the number of calls received in a second as a Poisson process with rate \(\lambda = 1\).

2. **Poisson Distribution:**
   - The Poisson distribution is given by:
     \[
     P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
     \]
   - Here, \(\lambda = 1\) and we need to find the probability that \(X \geq 5\).

3. **Calculating the Probability:**
   - We need to find \(P(X \geq 5)\):
     \[
     P(X \geq 5) = 1 - P(X < 5)
     \]
   - We calculate \(P(X < 5)\) by summing the probabilities from \(X = 0\) to \(X = 4\):
     \[
     P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
     \]
   - Using the Poisson formula:
     \[
     P(X = k) = \frac{1^k e^{-1}}{k!}
     \]
     \[
     P(X = 0) = \frac{1^0 e^{-1}}{0!} = e^{-1}
     \]
     \[
     P(X = 1) = \frac{1^1 e^{-1}}{1!} = e^{-1}
     \]
     \[
     P(X = 2) = \frac{1^2 e^{-1}}{2!} = \frac{e^{-1}}{2}
     \]
     \[
     P(X = 3) = \frac{1^3 e^{-1}}{3!} = \frac{e^{-1}}{6}
     \]
     \[
     P(X = 4) = \frac{1^4 e^{-1}}{4!} = \frac{e^{-1}}{24}
     \]
   - Summing these probabilities:
     \[
     P(X < 5) = e^{-1} + e^{-1} + \frac{e^{-1}}{2} + \frac{e^{-1}}{6} + \frac{e^{-1}}{24}
     \]
     \[
     P(X < 5) = e^{-1} \left(1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24}\right)
     \]
     \[
     P(X < 5) = e^{-1} \left(\frac{24}{24} + \frac{24}{24} + \frac{12}{24} + \frac{4}{24} + \frac{1}{24}\right)
     \]
     \[
     P(X < 5) = e^{-1} \left(\frac{65}{24}\right)
     \]
     \[
     P(X < 5) = \frac{65}{24} e^{-1}
     \]
   - Therefore:
     \[
     P(X \geq 5) = 1 - \frac{65}{24} e^{-1}
     \]

4. **Mean Interval Between Such Seconds:**
   - The mean interval of time between such seconds can be found using the reciprocal of the probability:
     \[
     \text{Mean interval} = \frac{1}{P(X \geq 5)}
     \]

The final answer is \( \boxed{ P(X \geq 5) = 1 - \frac{65}{24} e^{-1} } \) and the mean interval is \(\frac{1}{1 - \frac{65}{24} e^{-1}}\).