GeezerVison

This confuses me. Once __wakeup() executes, it proceeds without waiting to call __step() -- there's no scheduling involved that I can see in the source code (unless the .c extension is different). The call to future.result() immediately produces a value or raises -- it doesn't block (there's no await there). I do think it is possible to enter step 2b without going through 2a when __step() is scheduled directly by __init__(). So maybe it's clearer to say that we go through either 2a or 2b?

I may have got my wires crossed here, because it is all confusing. Initially, I thought __wakeup() was called directly (futures just calling registered callbacks) and _wakeup scheduling __step,.. But then it turns out that futures schedule their callbacks... I'll have another look.

Yes, you were right. So, there is no 2a or2b anymore just 2.
The main point I'm trying to get across here is that there can be a _fut_waiter present in a done state, i.e. the
mere presence of a _fut_waiter does not indicate that the Task is still blocked. I hope I'm not making this sound more complicated than it is.

the mere presence of a _fut_waiter does not indicate that the Task is still blocked

Yes, that's crucial information.