Explanation:
again, I assume the -1 exponent means "inverse function" and not 1/function.
the inverse function of g(x) :
g(x) = y = (x + 13)/5
5y = x + 13
5y - 13 = x
now we rename y to x and vice versa, so that we create a "normal" function :
g-¹(x) = y = 5x - 13
(g-¹ × g)(2) = (5×2 - 13)×(2 + 13)/5 =
= (10 - 13)×15/5 = -3×3 = -9
h-¹(4) means the x-value when h(x) = 4.
but there is no h(x) = 4.
undefined.