CREATE TABLE table_23235767_4 (year VARCHAR2, score_in_the_final VARCHAR2, surface VARCHAR2)

Name the number of year for 6 3, 6 2 hard surface
SELECT COUNT(year) FROM table_23235767_4 WHERE score_in_the_final = "6–3, 6–2" AND surface = "Hard"
SELECT COUNT("year") FROM "table_23235767_4" WHERE "6–3, 6–2" = "score_in_the_final" AND "Hard" = "surface"