To solve this, we need to calculate the probability of two events happening in sequence:

1. First, picking an 8:
   There is 1 card with an 8 in the set {2, 3, 4, 5, 6, 7, 8}, so the probability of picking an 8 is:
   $P(\text{8}) = \frac{1}{7}$

2. Second, picking a number greater than 4:
   The numbers greater than 4 in this set are {5, 6, 7, 8}, so there are 4 favorable outcomes. The probability of picking a number greater than 4 is:
   $P(\text{> 4}) = \frac{4}{7}$

Since these two events are independent (you replace the card after each pick), the total probability of both events happening is the product of the individual probabilities: $P(\text{8 and > 4}) = P(\text{8}) \times P(\text{> 4}) = \frac{1}{7} \times \frac{4}{7} = \frac{4}{49}$

Thus, the probability of picking an 8 and then picking a number greater than 4 is $\frac{4}{49}$.

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Would you like further details or have any other questions? Here's some related questions for you:

1. What is the probability of picking a 6 and then picking a number less than 6?
2. If there were two 8s in the deck, how would that affect the probability of picking an 8 first?
3. How would the probability change if there were 10 cards in total instead of 7?
4. What is the probability of picking two numbers greater than 4 consecutively?
5. What is the probability of picking a card smaller than 5 and then a card smaller than 3?

Tip: When dealing with independent events (like picking cards with replacement), you can always multiply the probabilities of each event occurring.