Given,
f(x)=x3+ax2+bx–12
Let us assume
x−2=0  and  x+3=0
⇒x=2  and  x=−3

Also given that, (x−2) and (x+3) are factors of f(x)
∴  By factor theorem, f(2)=0  and  f(−3)=0

Now, f(2)=0
⇒23+a(2)2+b(2)−12=0
⇒8+4a+2b−12=0
⇒4a+2b−4=0
Dividing te equation by 2
⇒2a+b−2=0...(i)

And, f(−3)=0
⇒(−3)3+a(−3)2+b(−3)−12=0
⇒−27+9a−3b−12=0
⇒9a−3b−39=0
Dividing the equation by 3
⇒3a–b=13...(ii)

Now, adding equation (i) and equation (ii) we get,
(2a+b)+(3a−b)=2+13
⇒2a+3a+b−b=15
⇒5a=15
⇒a=515​=3

Substituting this value of a in equation (i)
⇒2a+b=2
⇒2(3)+b=2
⇒6+b=2
⇒b=2−6
⇒b=−4

Hence, a=3  and  b=−4