(a) Showing \( f \) is even:

\( f(-x) = \arcsin\left(\frac{(-x)^2-1}{(-x)^2+1}\right) = \arcsin\left(\frac{x^2-1}{x^2+1}\right) = f(x) \)
Conclusion: \( f \) is even since \( f(-x) = f(x) \).

(b) Horizontal Asymptote:

\( \lim_{x\to\infty} \arcsin\left(\frac{x^2-1}{x^2+1}\right) = \arcsin\left(\lim_{x\to\infty}\frac{1-1/x^2}{1+1/x^2}\right) = \arcsin(1) = \frac{\pi}{2} \)
Asymptote: \( \boxed{y = \frac{\pi}{2}} \)

(c)(i) Derivative Calculation:

Let \( y = f(x) \), then \( \sin y = \frac{x^2-1}{x^2+1} \)
Differentiate implicitly: \( y’\cos y = \frac{4x}{(x^2+1)^2} \)
Using \( \cos y = \frac{2|x|}{x^2+1} \):
\( y’ = \frac{4x}{(x^2+1)^2} \cdot \frac{x^2+1}{2|x|} = \frac{2x}{|x|(x^2+1)} = \frac{2x}{\sqrt{x^2}(x^2+1)} \)
Result: \( \boxed{f'(x) = \frac{2x}{\sqrt{x^2}(x^2+1)}} \)

(c)(ii) Decreasing for \( x < 0 \):

For \( x < 0 \), \( \sqrt{x^2} = |x| = -x \), so:
\( f'(x) = \frac{2x}{-x(x^2+1)} = \frac{-2}{x^2+1} < 0 \)
Since derivative is negative, \( f \) is decreasing for \( x < 0 \).

(d) Inverse Function:

Let \( y = g(x) \), then:
\( \sin y = \frac{x^2-1}{x^2+1} \)
Solve for \( x \): \( x^2 = \frac{1+\sin y}{1-\sin y} \)
Since \( x \geq 0 \): \( x = \sqrt{\frac{1+\sin y}{1-\sin y}} \)
Thus: \( \boxed{g^{-1}(x) = \sqrt{\frac{1+\sin x}{1-\sin x}}} \)

(e) Domain of \( g^{-1} \):

The range of \( g \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right) \), so:
\( \boxed{\text{Domain of } g^{-1} = \left[-\frac{\pi}{2}, \frac{\pi}{2}\right)} \)

(f) Graph Sketch:

Key features:
– Vertical asymptote at \( x = \frac{\pi}{2} \)
– Passes through (0,1) since \( g^{-1}(0) = 1 \)
– Approaches infinity as \( x \to \frac{\pi}{2}^- \)
– Approaches 0 as \( x \to -\frac{\pi}{2}^+ \)
(Graph would be included here in actual implementation)