Everything within () in (1) is an independent variable. You can make the things in brackets equal to t, and then this t equals x in f(x). You can use f(x) to solve the domain first, then make t=x, and then you can solve the initial x, which is method of substitution.

(2) For example, the definition domain of f(x) is (-8,8) and the range is (-100, 100). It passed (-4,4). What point, domain and range does f(( 1/3)x) pass through?

Let ((1/3)x)=t, because their analytical formulas are the same, so the point where f(t) crosses (-4,4) has the domain of (-8,8) and the range of (-100, 100).

Because their analytical expressions are the same ((kloc-0//3) x) is defined as (-8,8).

The range of f(( 1/3)x) is (-100, 100).

It passes (-4,4), so ((1/3)x)=-4, x=- 12, and it passes (-12,4).

(3) Because the analytical formula is the same, but the independent variable is changed, you need to understand the process of changing the independent variable. For example, if X is changed to 2x, the domain will be reduced by 2 times. Don't worry, your teacher will talk about this later. It's blurred now. I wish I could experience it more.

I hope it helps you.