Thesql… compare and match items in same column

I suspect (with a high degree of of confidence) that the / isn't considered a "word" character. If that is the case, then Fine Tuning MySQL Full Text Search in the manual explains how to accomplish what you're wanting to do.

If you want to change the set of characters that are considered word characters, you can do so in several ways, as described in the following list. After making the modification, you must rebuild the indexes for each table that contains any FULLTEXT indexes. Suppose that you want to treat the hyphen character ('-') as a word character. Use one of these methods...

The second option offered there is probably the simplest, since you can apparently just edit the Character Definition Array in the appropriate XML file -- assuming you can use a one-byte character set such as latin1 for this column of this table. If you need utf8 or anything else multibyte, it apparently means recompiling the server from source after customizing the source to your specific needs.

I'm sure there are many ways to do this. Here's one:

CREATE TABLE QA (id int, question int, answer int, user int);

INSERT INTO QA
VALUES (1, 11, 2, 10)
      ,(2, 12, 2, 10)
      ,(3, 13, 3, 10)
      ,(4, 14, 4, 10)
      ,(5, 11, 2, 11)
      ,(6, 12, 2, 11)
      ,(7, 13, 4, 11)
      ,(8, 14, 1, 11)
      ,(9, 11, 2, 12)
      ,(10, 12, 2, 12)
      ,(11, 13, 1, 12)
      ,(12, 14, 1, 12)
;

SET @QCount := (SELECT COUNT(*) as QCount from QA WHERE QA.user = 10);
SET @QCount := IF(@QCount = 0, NULL, @QCount);

SELECT q1.user, ROUND(COUNT(a1.answer) * 100.0 / @QCount, 1) as percent
  FROM QA q1
         LEFT  JOIN QA a1 ON (    q1.question = a1.question
                              AND q1.answer = a1.answer
                              AND a1.user = 10
                             )
 WHERE q1.user <> 10
 GROUP BY q1.user
;

Tested via this db-fiddle

Here's how it works:

I am assuming that (in this case, at least) user 10's answers should be considered to be the answer key. We select the number of questions for that user to get the total question count.

Then, for each other user who has at least one answer, we get all the answers for each user, and LEFT JOIN them to the answer key questions and answers. If the answer key answer and the user's answer don't match, then the left join fails to find a match.

So, for all the right answers, we have a matched row from the answer key; for the wrong answers, we have the user's answer row, with no matched row.

We then count the number of matched rows for each user. The COUNT function doe snot count values that are NULL; by counting on a column from the LEFT JOIN table (the answer key), we'll only get a count of the right answers. NOTE: if, for some reason, there's no answer for some questions for a given those won't exists and won't be counted; so, they'll be treated as wrong answers.

To get the percentage of right answers, we simply multiply the number of wrong answers by 100, and divide by the total number of questions in the answer key. I leave the formatting aspect of showing an actual percent symbol up to the user, if required.

Note that the actual results are 50% for users 11 and 12; you showed 75% for one, bit the answers in your sample data show only two right for each user.