Depleted

Для MySQL 8.0 и выше:

WITH RECURSIVE `subs` (`name`, `value`, `order`) AS (
  SELECT CONCAT('{{> ', `p`.`name`, '}}'), `p`.`value`, ROW_NUMBER() OVER ()
    FROM `templates` AS `t`
    JOIN `parts` AS `p`
      ON `t`.`value` LIKE CONCAT('%{{> ', `p`.`name`, '}}%')
    WHERE `t`.`id` = 1
),
`result` (`value`, `order`) AS (
  SELECT `value`, 0
    FROM `templates`
    WHERE `id` = 1
  UNION SELECT REPLACE(`r`.`value`, `s`.`name`, `s`.`value`), `s`.`order`
    FROM `result` AS `r`
    JOIN `subs` AS `s`
      ON `s`.`order` = `r`.`order` + 1
)
SELECT `value`
  FROM `result`
  ORDER BY `order` DESC
  LIMIT 1

Для старых версий даже пробовать не буду, мне столько не выпить.

SELECT PeopleTable.gender, COUNT(*) AS GenderCount
  FROM ListTable 
    INNER JOIN PeopleTable ON ListTable.userID = PeopleTable.id
  GROUP BY PeopleTable.gender

А если нужен и ListTable, у которых нет соответствия в PeopleTable

SELECT ISNULL(PeopleTable.gender, '-') [Пол], COUNT(*) [К-во]
  FROM ListTable 
    LEFT JOIN PeopleTable ON ListTable.userID = PeopleTable.id
  GROUP BY ROLLUP (PeopleTable.gender)
UNION
SELECT ' ', COUNT(*)
  FROM ListTable   
ORDER BY 1 DESC, 2

Две последние строчки показывают общее количество. Если они не совпадают выполните проверку. Руководство.

DBCC CHECKTABLE ('PeopleTable')
DBCC CHECKTABLE ('ListTable')

Попробуйте:

SELECT 
	COUNT(DISTINCT CASE WHEN people.gender = 'M' THEN list.id END) AS mens,
	COUNT(DISTINCT CASE WHEN people.gender = 'F' THEN list.id END) AS womens,
	COUNT(DISTINCT CASE WHEN people.gender IS NULL THEN list.id END) AS unknown
FROM ListTable AS list 
LEFT JOIN PeopleTable AS people ON (list.userID = people.id) 
;

MS SQL Fiddle