We know that \(\displaystyle 4=2^2\) and \(\displaystyle 8=2^3\).

\(\displaystyle 4^{2x-1}=8^{4x+2}\Rightarrow \left [ 2^2 \right ]^{2x-1}=\left [ 2^3 \right ]^{4x+2}\)

\(\displaystyle \Rightarrow 2^{2(2x-1)}=2^{3(4x+2)}\)

Since the bases are the same, we can set the exponents equal to one another:

\(\displaystyle 2(2x-1)=3(4x+2)\Rightarrow 4x-2=12x+6\)

\(\displaystyle \Rightarrow 4x-12x=6+2\)

\(\displaystyle \Rightarrow -8x=8\)

\(\displaystyle \Rightarrow x=\frac{8}{-8}=-1\)

We know that any number raised to the power of zero is equal to \(\displaystyle 1\). Then we can substitute \(\displaystyle 1\) with \(\displaystyle 3^0\):

\(\displaystyle 3^0=1\Rightarrow3^{4x-2}=1=3^0\)

Now we need to solve the following equation:

\(\displaystyle 3^{4x-2}=3^0\)

Since the bases are the same, we can set the exponents equal to one another:

\(\displaystyle 4x-2=0\Rightarrow 4x=2\Rightarrow x=\frac{2}{4}=\frac{1}{2}=0.5\)

We know that any number raised to the power of zero is equal to \(\displaystyle 1\):

\(\displaystyle y^0=1\)

Then:

\(\displaystyle 7^{2x-3}=y^0=1\)

Again we know that any number raised to the power of zero is equal to \(\displaystyle 1\), so we can substitute \(\displaystyle 1\) with \(\displaystyle 7^0\):

\(\displaystyle 7^{2x-3}=1=7^0\Rightarrow 7^{2x-3}=7^0\)

Since the bases are the same, we can set the exponents equal to one another:

\(\displaystyle 2x-3=0\Rightarrow 2x=3\Rightarrow x=\frac{3}{2}=1.5\)

We know that \(\displaystyle 4=2^2\) and \(\displaystyle 8=2^3\).
 

\(\displaystyle 4^{\frac{x+2}{x-2}}=8\Rightarrow \left [ 2^2 \right ]^{\frac{x+2}{x-2}}=2^3\Rightarrow 2^{2(\frac{x+2}{x-2})}=2^3\)

Since the bases are the same, we can set the exponents equal to one another:

\(\displaystyle 2(\frac{x+2}{x-2})=3\Rightarrow (\frac{2x+4}{x-2})=3\)

\(\displaystyle \Rightarrow 2x+4=3(x-2)\Rightarrow 2x+4=3x-6\)

\(\displaystyle \Rightarrow 4+6=3x-2x\Rightarrow x=10\)

We know that any number raised to the power of zero is equal to \(\displaystyle 1\). Therefore we can substitute \(\displaystyle 1\) with \(\displaystyle 5^0\) in the equation:

\(\displaystyle 5^{(\frac{2x-8}{x-3})}=5^0\)

Since the bases are the same, we can set the exponents equal to one another:

\(\displaystyle {(\frac{2x-8}{x-3})}=0\Rightarrow 2x-8=0\Rightarrow 2x=8\Rightarrow x=\frac{8}{2}=4\)

We know that \(\displaystyle 27=3^3\):

\(\displaystyle 27\times 3^x=3^3\times 3^x\Rightarrow 27\times 3^x=3^{3+x}\)
 

\(\displaystyle 3^{\frac{x^2-5}{x-2}}=27\times 3^x\Rightarrow 3^{\frac{x^2-5}{x-2}}=3^{3+x}\)

Since the bases are the same, we can set the exponents equal to one another:

\(\displaystyle \frac{x^2-5}{x-2}=3+x\Rightarrow x^2-5=(x-2)(3+x)\)

\(\displaystyle \Rightarrow x^2-5=3\times x+x\times x-2\times 3-2\times x\)

\(\displaystyle \Rightarrow x^2-5=3x+x^2-6-2x\)

\(\displaystyle \Rightarrow x^2-5=x^2+x-6\)

\(\displaystyle \Rightarrow x^2-x^2-5+6=x\)

\(\displaystyle \Rightarrow x=1\)

\(\displaystyle \left ( f \circ g\right ) (0) = f(g(0))\)

First, evaluate \(\displaystyle g(0)\) by using the definition of \(\displaystyle g\) for nonnegative values of \(\displaystyle x\).

\(\displaystyle g(x) = x- 4\)

\(\displaystyle g(0) = 0- 4 = -4\)

Therefore, \(\displaystyle \huge \left ( f \circ g\right ) (0) = f(g(0)) = f (-4)\)

\(\displaystyle f(x) = x^{2}\), so

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

\(\displaystyle g (x) =8 - \frac{9}{10} x\)

\(\displaystyle g (3) =8 - \frac{9}{10} \cdot 3\)

\(\displaystyle g (3) =8 - \frac{27}{10}\)

\(\displaystyle g (3) = \frac{80 }{10}- \frac{27}{10}\)

\(\displaystyle g (3) = \frac{53 }{10}\)

\(\displaystyle g (3) = 5 \frac{3 }{10}\)

\(\displaystyle 7x - 17 >-5\)

\(\displaystyle 7x - 17 +17> -5+17\)

\(\displaystyle 7x >12\)

\(\displaystyle 7x\div 7 >12 \div 7\)

\(\displaystyle x > 1 \frac{5}{7}\)

\(\displaystyle \left ( f\circ g\right ) (x) = f(g(x)) = f (\sqrt{x+9})\)

Substitute \(\displaystyle \sqrt{x+9}\) for \(\displaystyle x\) in the definition of \(\displaystyle f\):

\(\displaystyle f (x) = x^{2} - 10\)

\(\displaystyle f (\sqrt{x+9} ) =\left (\sqrt{x+9} \right )^{2} - 10\)

\(\displaystyle f (\sqrt{x+9} ) =x+9- 10\)

\(\displaystyle f (\sqrt{x+9} ) =x-1\)

Therefore, 

\(\displaystyle \left ( f\circ g\right ) (x) =x-1\)

and

\(\displaystyle \left ( f\circ g\right ) (40) =40-1 = 39\)