使用相关的层次结构查询为每个工作日生成一行，然后对每一天的小时总数求和:

Use a correlated hierarchical query to generate one row for each work day and then sum the hours for each day:

SELECT task,
       COALESCE( SUM( end_time - start_time ), 0 ) * 24 AS total_hours
FROM   (
  SELECT task,
         GREATEST( t.start_time, d.column_value + INTERVAL '8' HOUR ) AS start_time,
         LEAST( t.end_time, d.column_value + INTERVAL '18' HOUR ) AS end_time
  FROM   your_table t
         LEFT OUTER JOIN
         TABLE(
           CAST(
             MULTISET(
               SELECT TRUNC( t.start_time + LEVEL - 1 )
               FROM   DUAL
               WHERE  TRUNC( t.start_time + LEVEL - 1 ) - TRUNC( t.start_time + LEVEL - 1, 'iw' ) < 6
               CONNECT BY TRUNC( t.start_time + LEVEL - 1 ) < t.end_time
             ) AS SYS.ODCIDATELIST
           )
         ) d
         ON (   t.end_time   > d.column_value + INTERVAL  '8' HOUR
            AND t.start_time < d.column_value + INTERVAL '18' HOUR )
)
GROUP BY task;