The problem asks for:

1. The domain, range, and graph of the function $f(x) = \sqrt{x^2 - 1}$ and $g(x) = \frac{2}{x}$.
2. The domain and composition of $f(g(x))$.

Let's break it down:

a) Domain, Range, and Graph of $f(x) = \sqrt{x^2 - 1}$

Domain:

For the square root function to be real, the expression inside the square root must be non-negative: $x^2 - 1 \geq 0$ Solving this inequality: $x^2 \geq 1 \quad \Rightarrow \quad |x| \geq 1$ So, the domain of $f(x)$ is: $x \in (-\infty, -1] \cup [1, \infty)$

Range:

Since $f(x) = \sqrt{x^2 - 1}$, the smallest value $x^2 - 1$ can take is $0$, and there is no upper bound. Therefore, the range is: $[0, \infty)$

Graph:

The graph of $f(x) = \sqrt{x^2 - 1}$ resembles a curve starting from $(1, 0)$ and $(-1, 0)$, and increases as $|x|$ increases.

b) Domain and Composition of $f(g(x))$

Domain of $g(x) = \frac{2}{x}$:

The function $g(x)$ is undefined at $x = 0$, so the domain is: $x \in (-\infty, 0) \cup (0, \infty)$

Composition $f(g(x))$:

For $f(g(x)) = f\left( \frac{2}{x} \right)$, we need to ensure that the argument of $f(x)$, which is $\frac{2}{x}$, satisfies the domain condition of $f(x)$.

We require: $\left( \frac{2}{x} \right)^2 - 1 \geq 0$ Simplifying: