Pre-Calculus question, can anyone help me out

1.

Evaluate h(4), where h=g o f.
f(x)=3(index)√(x^2-6), g(x)=7x^3+6
h(4)=?

I'm so confused with index.

2.

Find f o g o h.
f(x)=√(x-3)
g(x)=x^2
h(x)=x^3+9

h(x) =>
g(f(x)) =>
7 * ((x^2 - 6)^(1/3))^3 + 6 =>
7 * (x^2 - 6)^((1/3) * 3) + 6 =>
7 * (x^2 - 6) + 6 =>
7x^2 - 42 + 6 =>
7x^2 - 36

h(4) =>
7 * (4)^2 - 36 =>
7 * 16 - 36 =>
112 - 36 =>
76



f(g(h(x)))

First, find g(h(x))

g(h(x)) => h(x)^2 = (x^3 + 9)^2 = x^6 + 18x^3 + 81

f(g(h(x))) =>
f(x^6 + 18x^3 + 81) =>
sqrt((x^6 + 18x^3 + 81) - 3) =>
sqrt(x^6 + 18x^3 + 78)

index - sometimes used instead of power. so,
3(index)√(x^2-6) means 3^√(x^2-6)

g o f. - simply means g(f(x) so,
(g * f)(x) = g(f(x)) = g(3(index)√(x^2-6)
since, g(x) = 7x^3+6 then substitute the value of x in f(x) by 3(index)√(x^2-6)
g(f(x))= 7[(3(index)√(x^2-6)]^3 + 6 = 7[3^[(x^2-6)^(3/2] + 6
=7(3^(x^3-6√6) + 6 then
if x = 4, that is h(4) ill be:
h(4) = g o f = 7(3^((4)^3-6√6) + 6

Work:
7[(3(index)√(x^2-6)]^3
=7[3^)√(x^2-6)]^3
=7{3^[(x^2-6)^[(1/2)*3]}
=7[3^[(x^2-6)^(3/2]

Find f o g o h.
f(x)=√(x-3)
g(x)=x^2
h(x)=x^3+9

(f o g o h)(x) = f(g(h(x)))
given: h(x) = x^3+9
= f(g(x^3+9))
substitute the value of x in g(x) by x^3+9:
= f(x^3+9)^2)
substitute the value of x in f(x) by x^3+9:
= √[(x^3+9)^2)-3]