A function is odd if and only if $\displaystyle f(-x) = -f(x)$ for each value of $\displaystyle x$ in the domain; it is even if and only if $\displaystyle f(-x) = f(x)$ for each value of $\displaystyle x$ in the domain. To disprove a function is odd or even, we need only find one value of $\displaystyle x$ for which the appropriate statement fails to hold. 

Consider $\displaystyle x = 1$:

$\displaystyle f(x) = x^{3} - 2$

$\displaystyle f(1) = 1^{3} - 2 = 1-2 = -1$

$\displaystyle f(x) = x^{3} - 2$

$\displaystyle f(-1) = (-1)^{3} - 2 = -1 -2 = -3$

$\displaystyle f(-1 ) \neq - f(1)$, so $\displaystyle f$ is not an odd function; $\displaystyle f(-1 ) \neq f(1)$, so $\displaystyle f$ is not an even function.

$\displaystyle (f \circ g) (-4) = f (g(-4))$

First we evaluate $\displaystyle g (-4)$. Since the parameter is negative, we use the first half of the definition of $\displaystyle g$:

$\displaystyle g (x) = 2x-5$

$\displaystyle g (-4) = 2 \left ( -4\right )-5 = -8 - 5 = -13$

$\displaystyle f (g(-4)) = f(-13)$; since the parameter here is again negative, we use the first half of the definition of $\displaystyle f$:

$\displaystyle f(-13) = 1$

Therefore, $\displaystyle (f \circ g) (-4) = 1$.

$\displaystyle f (x) = \left \lfloor x ^{2} \right \rfloor - \left \lfloor 4 x \right \rfloor + 8$

$\displaystyle f \left ( \frac{3}{5} \right ) = \left \lfloor \left (\frac{3}{5} \right ) ^{2} \right \rfloor - \left \lfloor 4 \cdot \frac{3}{5}\right \rfloor + 8$

$\displaystyle = \left \lfloor \frac{9}{25} \right \rfloor - \left \lfloor \frac{12}{5}\right \rfloor + 8$

$\displaystyle = \left \lfloor \frac{9}{25} \right \rfloor - \left \lfloor 2 \frac{2}{5}\right \rfloor + 8$

$\displaystyle =0 - 2 + 8 = 6$

We start by finding g(2):

$\displaystyle g(2)=3\cdot2-2 = 6-2 = 4$

Then we find f(g(2)) which is f(4): 

$\displaystyle f(4)=4^{2}-\sqrt{4}=16-2=14$

$\displaystyle \left ( q \circ p\right )(x) = q(p(x))= q(x^{2})$ by definition. $\displaystyle q$ is piecewise defined, with one defintion for negative values of the domain and one for nonnegative values. However, $\displaystyle p(x)= x^{2}$ is nonnegative for all real numbers, so the defintion for nonnegative numbers, $\displaystyle q(x )= x-1$, is the one that will always be used. Therefore,

$\displaystyle \left ( q \circ p\right )(x) = q(p(x))= q(x^{2}) = x^{2}-1$ for all values of $\displaystyle x$.

$\displaystyle \left ( q \circ p\right )(x) = q(p(x))= q( 9 - x^{2})$ by definition.

 $\displaystyle q$ is piecewise defined, with one defintion for negative values of the domain and one for nonnegative values. 

If $\displaystyle 9 -x^{2} < 0$, then we use the definition $\displaystyle q(x)= 0$. This happens if

$\displaystyle 9 -x^{2} < 0$

$\displaystyle 9 < x^{2}$

$\displaystyle x > 3$ or $\displaystyle x < -3$

Therefore, the defintion of  $\displaystyle \left ( q \circ p\right )(x)$ for $\displaystyle x > 3$ or $\displaystyle x < -3$ is

$\displaystyle \left ( q \circ p\right )(x) = 0$

Subsquently, if $\displaystyle -3 \le x \le 3$, we use the defintion $\displaystyle q(x) = x$, since $\displaystyle 9 - x^{2} \ge 0$:

$\displaystyle \left ( q \circ p\right )(x) =q( 9 - x^{2}) = 9 -x^{2}$.

The correct choice is

$\displaystyle \left ( q \circ p\right )(x) = \left\{\begin{matrix} 0& x < -3 \\ 9 - x^{2}& -3 \le x \le 3 \\ 0& x > 3 \end{matrix}\right.$

This can be understood better by substituting $\displaystyle t = \sqrt{x}$, and, subsequently, $\displaystyle t^{2} =\left ( \sqrt{x} \right )^{2} = x$ in the function's definition.

$\displaystyle x + 4 \sqrt{x}- 6 = t^{2} +4 t - 6$

which is now in standard quadratic form in terms of $\displaystyle t$.

Write this in vertex form by completing the square:

$\displaystyle t^{2} +4 t - 6 = t^{2}+4 t + \left ( \frac{4}{2} \right ) ^{2} - 6 - \left ( \frac{4}{2} \right ) ^{2}$

$\displaystyle = t^{2}+4 t +4- 6 - 4$

$\displaystyle =\left ( t+2 \right )^{2}- 10$

Substitute $\displaystyle \sqrt{x}$ back for $\displaystyle t$, and the original function can be rewritten as

$\displaystyle f(x)=\left ( \sqrt{x}+2 \right )^{2}- 10$.

To find the range, note that $\displaystyle \sqrt{x} \ge 0$. Therefore, 

$\displaystyle \sqrt{x} + 2 \ge 2$

$\displaystyle \left ( \sqrt{x}+2 \right )^{2} \ge 4$

and 

$\displaystyle f(x)=\left ( \sqrt{x}+2 \right )^{2}- 10 \ge -6$

The range of $\displaystyle f$ is the set $\displaystyle \left [ -6, \infty \right )$.

This can be understood better by substituting $\displaystyle t = x^{\frac{1}{3}}$, and, subsequently, $\displaystyle t^{2} =\left ( x^{\frac{1}{3}} \right )^{2} = x^{\frac{2}{3}}$ in the function's definition.

$\displaystyle x^{\frac{2}{3}} + 6x^{\frac{1}{3}}+6 = t^{2}+ 6t + 6$

which is now in standard quadratic form in terms of $\displaystyle t$.

Write this in vertex form by completing the square:

$\displaystyle t^{2}+ 6t + 6$

$\displaystyle = t^{2}+ 6t +\left ( \frac{6}{2} \right )^{2}+ 6 - \left ( \frac{6}{2} \right )^{2}$

$\displaystyle = t^{2}+ 6t +9+ 6 - 9$

$\displaystyle = (t+3)^{2} - 3$

Substitute $\displaystyle x^{\frac{1}{3}}$ back for $\displaystyle t$. The original function can be rewritten as

$\displaystyle f(x)= (x^{\frac{1}{3}}+3)^{2} - 3$

or, in radical form,

$\displaystyle f(x)= (\sqrt[3]{x }+3)^{2} - 3$

$\displaystyle \sqrt[3]{x}$ can assume any real value; so, subsequently, can $\displaystyle \sqrt[3]{x} + 3$. But its square must be nonnegative, so

$\displaystyle \left (\sqrt[3]{x} + 3 \right )^{2} \ge 0$

and 

$\displaystyle f(x)= (\sqrt[3]{x }+3)^{2} - 3 \ge -3$

The range of $\displaystyle f$ is $\displaystyle [-3, \infty)$

In order to find exactly the $\displaystyle a,b$ values where the equations intersect and when $\displaystyle f(g(0))=1$. We need to consider each piece of information seperately.

Let's start with $\displaystyle f(g(0))=1$. Plugging $\displaystyle g$ into $\displaystyle f$, we have $\displaystyle f(g(x)) = 2(-x^2-b)^2+a$. Plugging 0 into this, we have$\displaystyle f(g(0))=2b^2+a$. This in turn equals 1, because we were given that piece of information in the beginning. So we end up with $\displaystyle 2b^2+a=1$

Now let's shift our attention to "intersect only when $\displaystyle x=1$" That means, if we plug 1 into both equations, we can set them equal to each other.

$\displaystyle 2(1)^2+a=-(1)^2-b$ becomes $\displaystyle 2+a = -1-b$ becomes $\displaystyle a+b= -3$.

Now we have two different equations arising from the two previous paragraphs.

$\displaystyle a+b = -3$

$\displaystyle a+2b^2=1$

We can solve this system of equations using the substitution method.

Solving for $\displaystyle a$ in the first equation gives $\displaystyle a=-3-b$.

Plugging this equation in for $\displaystyle a$ the 2nd equation gives $\displaystyle (-3-b) +2b^2 =1$. Using algebra on this equation we get $\displaystyle 2b^2-b-4=0$

Now we are ready to use the quadratic formula to solve for $\displaystyle b$.

$\displaystyle b = \frac{1\pm \sqrt{1^2-(4)(2)(-4)}}{2(2)}=\frac{1}{4}(1\pm \sqrt{33})$

Finally, since we're told in the beginning that $\displaystyle b>0$, we must pick the plus sign in our solution for $\displaystyle b$. Hence

$\displaystyle b =\frac{1}{4}(1+ \sqrt{33})$.

This question is asking us to find the composition of f and g. In order to do this we need to plug g(x) into the x value in f(x).

$\displaystyle \left ( f \circ g\right )(x)$

$\displaystyle = f(g(x))$

$\displaystyle = f \left ( \frac{1}{2x} \right )$

$\displaystyle = \left ( \frac{1}{2x} \right )^{2} - 6$

$\displaystyle = \frac{1}{4x^{2}} - 6$

$\displaystyle = \frac{1}{4x^{2}} - \frac{24x^{2}}{4x^{2}}$

$\displaystyle = \frac{1 - 24x^{2}}{4x^{2}}$