We have $f(f(f(x))) = a + b(a+b(a+bx)) = a + ab + b^2(a+bx) = a + ab + ab^2 + b^3x$. Then
f(f(f(1)))=a+ab+ab2+b3=14f(f(f(0)))=a+ab+ab2=−13
So $b^3 = 14+13 = 27$, that is, $b=3$. Substituting $b$ in the second equation gives $a+3a+9a = -13$, that is, $a = -1$.