To begin, note that lim x→-2 (x^2+x-2) = 0, so we want to have lim x→-2 (3x^2+ax+a+3) = 0, because then we can use L'Hopital's Rule and otherwise the limit does not exist.
As 3x^2+ax+a+3 is always continuous for any value of x, we need to find values of a such that 3x^2+ax+a+3 = 0 at x = -2. Hence 15-a=0, so a=15.
Lastly, we calculate the limit. lim x→-2 (3x^2+15x+18)/(x^2+x-2) = (using L'Hopital's) lim x→-2 (6x+15)/(2x+1) = -1.