\begin{align*} \text{Average rate of change}_{[-1,2]} &= \frac{f(2)-f(-1)}{2-(-1)}\\ &=\frac{(2+2^2)-(-1+2^{-1})}{3}\\ &=\frac{6-\Big(-1+\frac{1}{2}\Big)}{3}\\ &=\frac{6.5}{3}=2.166 \end{align*}

The rate of change of the average rate of change of a quadratic function is constant for equal length intervals. For this problem, the avg rate of change is increasing by 4 for each interval of 2. Therefore the average rate of change of $f$ on the interval $2\leq x \leq 4$ is 10.

Identify points which may or not be part of the domain. We have that expression within square root can not be negative so $x\geq 0$. Also denominator of a fraction can not be 0 so $x-3 \neq0 \Rightarrow x\neq 3$. We conclude that the domain of this function is any real number $x$ such that $x \geq 0$ and $x\neq 3$.

The function $f$ is an odd function if $f(-x) = -f(x)$ for all $x$ in the domain of $f$. The graph of an odd function is symmetric with respect to the origin.\\ For this problem: $f(-x)=\frac{1}{2}(2^{-x}-2^{-(-x)})=\frac{1}{2}(2^{-x}-2^x)=-\frac{1}{2}(2^x-2^{-x})=-f(x)$, thus $f$ is an odd function.

1. Vertical asymptotes of a function corespondent with denominator equal to 0 after the function is in the simplest form, therefore $\sqrt{x^2-9}=0$ yields two vertical asymptotes $x=3$ and $x=-3$
2. Horizontal asymptotes of a function describe the end behavior of a graph when $x \to \infty$ or when $x\to -\infty$ or $\lim\limits_{x\to \pm \infty}f(x)$. Thus, \begin{align*} \lim\limits_{x\to \pm \infty} \frac{x+1}{\sqrt{x^2-9}} &= \lim\limits_{x\to \pm \infty}\Big(\frac{x\Big(1+\frac{1}{x}\Big)}{\sqrt{x^2(1-\frac{9}{x^2}\Big)}}\Big)\\ &= \lim\limits_{x\to \pm \infty}\Big(\frac{x\Big(1+\frac{1}{x}\Big)}{|x|\sqrt{1-\frac{9}{x^2}}}\Big)\\ &\text{Note that } \lim\limits_{x\to \pm \infty}\frac{1}{x}=0 \text{ and } \lim\limits_{x\to \pm \infty} \frac{9}{x^2}=0\\ &= \lim\limits_{x\to \pm \infty} \frac{x}{|x|}\\ &=\lim\limits_{x\to \pm \infty} \pm 1\\ &=\pm 1 \end{align*} Hence, there are two horizontal asymptotes, but no oblique asymptotes, therefore in total 4 asympttotes.

It is given that $f(x)\geq 0$ for all $x$ and $f(2)=0$ which implies that $f$ is not crossing $x$-axis at $x=2$ rather bounces back, therefore function $f$ has a double root at $x=2$.
In this case function $f(x)$ will look like $f(x)=a(x-2)^2$. Using the other given $f(3)=3$ we can replace 3 for $x$ and find that $a(3-2)^2=3$ which yield that $a=3$. Now function $f$ is $f(x)=3(x-2)^2$. Finally $f(5)=3(5-2)^2=27$

In order to determine domain of $f\circ g$ we determine values to exclude from the domain of $(f \circ g)(x)$. If $x$ is not in the domain of $g$, it must not be in the domain of $f\circ g$. Zero 0 is not in the domain of $g(x)=\frac{3}{x}$, thus, 0 must be excluded from the domain of $f\circ g$.
Any $x$ for which $g(x)$ is not in the domain of $f$ must not be in the domain of $f\circ g$. $f(g(x))=\frac{2}{g(x)-1}$, we must exclude from the domain of $(f\circ g)$ any $x$ for which $g(x)=1$, or $\frac{3}{x}=1$ or $x=3$. The domain of $(f\circ g)$ is $$(-\infty, 0)\cup (0,3)\cup (3, \infty)$$

Replace $f(x)$ with $y$ and interchange the variables $x$ and $y$ to obtain $$x=\frac{2y}{3y+1}$$ Solve for $y$ \begin{align*} x&=\frac{2y}{3y+1}\\ x(3y+1) &= 2y &&\text{Cross multiply}\\ 3xy+x &= 2y &&\text{Use the Distributive Property}\\ 3xy-2y&=-x &&\text{Subtract }2y \text{ from both sides and subtract }x \text{ from both sides}\\ y(3x-2)&=-x &&\text{Factor}\\ y&=\frac{-x}{3x-2}&&\text{Divide by }3x-2\\ \end{align*} The inverse function is $$f^{-1}(x)=\frac{-x}{3x-2}$$ Check that $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$
$\displaystyle f^{-1}(f(x))=f^{-1}\Big(\frac{2x}{3x+1}\Big)=\frac{-\frac{2x}{3x+1}}{3\cdot \frac{2x}{3x+1}-2}=\frac{-2x}{3x+1}\cdot \frac{3x+1}{6x-6x-2}=\frac{-2x}{-2}=x$
$\displaystyle f(f^{-1}(x))=f\Big(\frac{-x}{3x-2}\Big)=\frac{2 \cdot \frac{-x}{3x-2}}{3 \cdot \frac{-x}{3x-2}+1}=\frac{-2x}{3x-2}\cdot \frac{3x-2}{-3x+3x-2}=\frac{-2x}{-2}=x$

If a function $f$ is one-to-one, then it has an inverse function $f^{-1}$. Domain of function $f$ is the Range of $f^{-1}$ and the Range of function $f$ is the Domain of $f^{-1}$.
For this problem the domain of $f^{-1}(x)$ is $[4, \infty)$ and the range of $f^{-1}(x)$ is $[1, \infty)$.