The equation given represents a horizontal line.  This means that every y-value on the domain is equal to $\displaystyle \frac{1}{5}$.

The answer is:  $\displaystyle [\frac{1}{5}]$

The x-intercept is the value of x that will result in the y value equalling zero and the y-intercept is the y value that results when the x equals zero.

Therefore we replace x with zero to find the y-intecerpt.

$\displaystyle y=2(0)-8$

$\displaystyle y=-8$

Now we replace the y with a zero and find our x-intercept

$\displaystyle 0=2x-8$

$\displaystyle 8=2x$

$\displaystyle 4=x$

To find the x and y intercept of the function we need to do algebraic opperations first and then plug in zero for x to find the y-intercept and plug in zero for y to find the x-intercept.

$\displaystyle y-4=x+8$

$\displaystyle y=x+12$

y-intercept

$\displaystyle y=0+12=12$

x-intercept

$\displaystyle 0=x+12$

$\displaystyle -12=x$

Evaluate $\displaystyle g(0)$:

$\displaystyle g(x)=5- \log _{4}\left ( x+5 \right )$

$\displaystyle g(0)=5- \log _{4}\left ( 0+5 \right )$

$\displaystyle =5- \log _{4} 5$

$\displaystyle =5- \frac{\ln 5}{\ln 4}$

$\displaystyle \approx 5- \frac{1.6094}{1.3863}$

$\displaystyle \approx 5- 1.16$

$\displaystyle \approx 3.84$

Set $\displaystyle f(x) = 0$ and solve for $\displaystyle x$:

$\displaystyle f(x) = 2 \cdot 3^{x-4}- 7$

$\displaystyle 2 \cdot 3^{x-4}- 7= 0$

$\displaystyle 2 \cdot 3^{x-4}=7$

$\displaystyle 3^{x-4}=\frac{7}{2} = 3.5$

$\displaystyle \ln \left (3^{x-4} \right )= \ln 3.5$

$\displaystyle (x-4 ) \ln 3 = \ln 3.5$

$\displaystyle x-4 = \frac{\ln 3.5}{ \ln 3} \approx \frac{1.2528}{1.0986} \approx 1.14$

$\displaystyle x\approx 5.14$

Evaluate $\displaystyle g(0)$:

$\displaystyle g(x) = 3 + \sqrt[4]{3x-5}$

$\displaystyle g(0) = 3 + \sqrt[4]{3\cdot 0-5}$

$\displaystyle g(0) = 3 + \sqrt[4]{ -5}$

A negative number does not have a real even-numbered root, so $\displaystyle g(0)$ is not a real number. Therefore, the graph of $\displaystyle g$ has no $\displaystyle y$-intercept.

Evaluate $\displaystyle f(0)$:

$\displaystyle f(x) = 4 + \sqrt[5]{2x-9}$

$\displaystyle f(0) = 4 + \sqrt[5]{2 \cdot 0 -9}$

$\displaystyle = 4 + \sqrt[5]{ -9}$

$\displaystyle \approx 4 + (-1.55)$

$\displaystyle \approx 2.45$

Evaluate $\displaystyle h(0)$:

$\displaystyle h(x) = 12 + \sqrt[3]{3x-7}$

$\displaystyle h(0) = 12 + \sqrt[3]{3\cdot 0 -7}$

$\displaystyle = 12 + \sqrt[3]{ -7}$

$\displaystyle \approx 12 +(-1.91)$

$\displaystyle \approx 10.09$

Set $\displaystyle h(x) = 0$ and solve for $\displaystyle x$:

$\displaystyle 6- \log _{3}\left ( x-7 \right )=0$

$\displaystyle \log _{3}\left ( x-7 \right )=6$

$\displaystyle x-7 = 3^{6}$

$\displaystyle x-7=729$

$\displaystyle x = 736$

Set $\displaystyle h(x) = 0$ and solve for $\displaystyle x$:

$\displaystyle 4+ \sqrt[3]{3x-7} = 0$

$\displaystyle \sqrt[3]{3x-7} = -4$

$\displaystyle \left (\sqrt[3]{3x-7} \right )^{3}= \left (-4 \right )^{3}$

$\displaystyle 3x-7 = -64$

$\displaystyle 3x= -57$

$\displaystyle x = -19$