Нет, вы не можете ссылаться на псевдоним столбца в том же запросе. Вам нужно будет сделать это подзапросом.
SELECT id, answerA, answerB, answerC, (answerA + answerM + answerD) AS total
FROM (
select
t1.id,
sum(t2.answ = t1.answ) as answerA,
sum(t3.answ = t1.answ) as answerM,
sum(t4.answ = t1.answ) as answerD,
sum(answerA + answerM + answerD) as total
from t1
left join t2 on t2.id = t1.t2fk
left join t3 on t3.id = t1.t3fk
left join t4 on t4.id = t1.t4fk
GROUP BY t1.id
) AS subquery
или вы можете написать эквивалентное выражение.
select
t1.id,
sum(t2.answ = t1.answ) as answerA,
sum(t3.answ = t1.answ) as answerM,
sum(t4.answ = t1.answ) as answerD,
sum(t1.answ IN (t2.answ, t3.answ, t4.answ)) as total
from t1
left join t2 on t2.id = t1.t2fk
left join t3 on t3.id = t1.t3fk
left join t4 on t4.id = t1.t4fk
GROUP BY t1.id