The scheduler that ran the least urgent task

A task scheduler is supposed to run tasks highest-priority first. Instead it runs them lowest-first, and the moment two tasks share a priority it crashes with TypeError: '<' not supported between instances of 'Task' and 'Task'. The heap operations themselves are used correctly. Find both bugs.

1import heapq
2 
3 
4class Task:
5    def __init__(self, name, priority):
6        self.name = name
7        self.priority = priority
8 
9 
10def run_order(tasks):
11    """Names in execution order: highest priority first."""
12    pq = []
13    for t in tasks:
14        heapq.heappush(pq, (t.priority, t))
15    order = []
16    while pq:
17        _, t = heapq.heappop(pq)
18        order.append(t.name)
19    return order
20 
21 
22print(run_order([Task("deploy", 3), Task("lint", 1), Task("test", 2)]))
23# ['lint', 'test', 'deploy'] — expected ['deploy', 'test', 'lint']
24 
25print(run_order([Task("a", 1), Task("b", 1)]))
26# TypeError: '<' not supported between instances of 'Task' and 'Task'

1. Is heapq a min-heap or a max-heap? What does heappop return, and what does that mean for "highest priority first"?
2. Explain the TypeError: how does Python compare the tuples (1, task_a) and (1, task_b), and why does the first test case never hit this?
3. Fix both problems with the standard idiom: negation plus a tie-breaking counter. Why must the counter sit *between* the priority and the task?
4. What extra property does the counter give you among equal-priority tasks?
5. Priorities are numbers here. What if they were strings ("high"/"low"), where negation is impossible? Give two options.
6. State the complexity of run_order.