Первый пришедший в голову вариант:

SELECT user_id
FROM some_table 
WHERE num IN (878, 925, 242)
GROUP BY user_id
HAVING COUNT(DISTINCT num) = 3 -- 3 количество вариантов
;

MySQL fiddle

Вариант с JOIN:

SELECT some_table.user_id
FROM some_table 
JOIN some_table st1 ON st1.user_id = some_table.user_id AND st1.num = 925
JOIN some_table st2 ON st2.user_id = some_table.user_id AND st2.num = 242
WHERE some_table.num = 878
;

MariaDB fiddle

Вариант с GROUP_CONCAT (номера должны быть отсортированы):

SELECT user_id, GROUP_CONCAT(DISTINCT num ORDER BY num)
FROM some_table 
WHERE num IN (878,925,242)
GROUP BY user_id
HAVING GROUP_CONCAT(DISTINCT num) = '242,878,925'
;

MySQL 8.0 Fiddle

Могу посоветовать следующую структуру:

create table `calendar` (
  `event` varchar(64),
  `year` smallint unsigned,
  `month` tinyint unsigned,
  `day` tinyint unsigned,
  `hour` tinyint unsigned,
  `minute` tinyint unsigned
);

insert into calendar values
('December 31 every year at 23:55', null, 12, 31, 23, 55),
('every minute in december 2020', 2020, 12, null, null, null),
('every hour at first of month', null, null, 1, null, 0)
;

select `event`
from `calendar`
where 
	(`year` is null or `year` = year(now())) and 
	(`month` is null or `month` = month(now())) and
	(`day` is null or `day` = day(now())) and
	(`hour` is null or `hour` = hour(now())) and
	(`minute` is null or `minute` = day(now()))
;

Запустить SQL запрос

Можно использовать оператор FIND_IN_SET:

SET @t = '528223,528224';

SELECT * FROM t 
WHERE FIND_IN_SET (t.id, @t );

MySQL fiddle

Учтите, значения в сете должны быть без пробелов.