Exercise 3 solution

    from datetime import date, timedelta

    def next_weekday(d, target):
        """Next date strictly after d that falls on weekday number target
        (Monday is 0, Sunday is 6)."""
        days_ahead = (target - d.weekday()) % 7
        if days_ahead == 0:
            days_ahead = 7
        return d + timedelta(days=days_ahead)

    d = date(2026, 11, 2)                  # a Monday
    print(next_weekday(d, 0))              # 2026-11-09
    print(next_weekday(d, 4))              # 2026-11-06
    print(next_weekday(d, 6))              # 2026-11-08
    print(next_weekday(d, 4).strftime("%A"))  # Friday

(All lines were run and printed exactly the values shown.)

Explanation: weekday() numbers Monday as 0 up to Sunday as 6. The modulo
7 turns "how many days forward" into a number from 0 to 6 even when the
target day is earlier in the week than today. A result of 0 would mean
"today", so it is replaced by 7 to get the following week, which is what
"next" means here.