fix(cron): prevent parallel job result loss on exception

Replace generator-based result collection with explicit per-future
handling. Each future is now processed independently with a 600s timeout.

Before: _results.extend(f.result() for f in _futures)
- One exception stops the generator, remaining results are lost
- No timeout: one hung job blocks the entire tick

After: as_completed() + per-future try/except
- Each future handled independently
- 600s timeout prevents indefinite blocking
- Failed futures are logged and counted as failures

cron/scheduler.py

@@ -1802,7 +1802,12 @@ def tick(verbose: bool = True, adapters=None, loop=None) -> int:
                 for job in parallel_jobs:
                     _ctx = contextvars.copy_context()
                     _futures.append(_tick_pool.submit(_ctx.run, _process_job, job))
-                _results.extend(f.result() for f in _futures)
+                for f in concurrent.futures.as_completed(_futures, timeout=600):
+                    try:
+                        _results.append(f.result())
+                    except Exception as exc:
+                        logger.error("Parallel cron job future failed: %s", exc)
+                        _results.append(False)
 
         # Best-effort sweep of MCP stdio subprocesses that survived their
         # session teardown during this tick.  Runs AFTER every job has