$\displaystyle 3t = 7u$

$\displaystyle \frac{3t}{3} =\frac{ 7u}{3}$

$\displaystyle t =\frac{ 7}{3}u$

Similarly,

$\displaystyle 4u = 7v$

$\displaystyle u = \frac{7}{4}v$

$\displaystyle 9v = 2w$

$\displaystyle v = \frac{2}{9}w$

By substitution:

$\displaystyle t =\frac{ 7}{3}u =\frac{ 7}{3} \cdot \frac{7}{4}v =\frac{ 7}{3} \cdot \frac{7}{4} \cdot \frac{2}{9}w$

$\displaystyle t = \frac{49}{54}w$

$\displaystyle 54 \cdot t =54 \cdot \frac{49}{54}w$

$\displaystyle 54t = 49 w$

$\displaystyle 8 ^{3x- 8} = 4 ^{y}$

$\displaystyle \left (2^{3} \right )^{3x- 8} = (2^{2}) ^{y}$

$\displaystyle 2^{3(3x- 8)} = 2^{2y}$

$\displaystyle 2y = 3(3x-8)$

$\displaystyle 2y = 9x-24$

$\displaystyle \frac{2y }{2}= \frac{9x-24}{2}$

$\displaystyle y = \frac{9}{2}x- 12$

$\displaystyle 6 ^{x} \cdot 36 ^{y} = 216 ^{t}$

$\displaystyle 6 ^{x} \cdot\left ( 6 ^{2} \right )^{y} = \left (6^{3} \right )^{t}$

$\displaystyle 6 ^{x} \cdot 6 ^{2 y} = 6^{3 t}$

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

$\displaystyle 3t = x+2y$

$\displaystyle \frac{3t }{3}= \frac{x+2y}{3}$

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

$\displaystyle \frac{4 ^{x} }{8 ^{y} } = 16^{t}$

$\displaystyle \frac{\left (2^{2} \right )^{x} }{\left (2^{3} \right ) ^{y} } =\left ( 2^{4} \right ) ^{t}$

$\displaystyle \frac{ 2^{2x} }{ 2^{3y} } = 2^{4t}$

$\displaystyle 2^{2x -3y} = 2^{4t}$

$\displaystyle 4t = 2x-3y$

$\displaystyle \frac{4t }{4}= \frac{2x-3y}{4}$

$\displaystyle t = \frac{1}{2}x- \frac{3}{4}y$

$\displaystyle (g \circ f) (-3) = g [f(-3)]$

From the diagram below, it can be seen that $\displaystyle f(-3) = 2$.

$\displaystyle (g \circ f) (-3) = g [f(-3)] = g(2)$

$\displaystyle g(x) = 4x+8$, so

$\displaystyle g(x) = 4 (2)+8 = 8 + 8 = 16$

Therefore, 

$\displaystyle (g \circ f) (-3) = 16$.

$\displaystyle \left (f \circ g \right ) (-3) = f \left [ g (-3)\right ]$

$\displaystyle g(x) = 4x+8$

$\displaystyle g(-3) = 4 (-3)+8 = -12 + 8 = -4$

$\displaystyle f \left [ g (-3)\right ] = f(-4)$.

Examine the diagram below.

As can be seen, $\displaystyle f(-4)= 4$. Therefore, $\displaystyle (f \circ g) (-3) = 4$.

The $\displaystyle y$-intercept of a function is the point at which $\displaystyle x = 0$, so we can find this by evaluating $\displaystyle g(0)$.

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

$\displaystyle g(0) = f(-0) = f(0)$

As seen in the diagram below, $\displaystyle f(0) = 5$.

Therefore, $\displaystyle g(0) = f(0) = 5$, and the $\displaystyle y$-intercept of the graph of $\displaystyle g$ is $\displaystyle (0, 5)$.

The $\displaystyle y$-intercept of a function is the point at which $\displaystyle x = 0$, so we can find this by evaluating $\displaystyle g(0)$.

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

$\displaystyle g(0) = -f(0)$

As seen in the diagram below, $\displaystyle f(0) = 5$.

Therefore, $\displaystyle g(0) = -f(0) = -5$, and the $\displaystyle y$-intercept of the graph of $\displaystyle g$ is $\displaystyle (0, -5)$

The $\displaystyle y$-intercept of a function is the point at which $\displaystyle x = 0$, so we can find this by evaluating $\displaystyle g(0)$.

$\displaystyle g(x) =- f^{-1}(x)$

$\displaystyle g(0) = - f^{-1}(0)$

The graph of $\displaystyle f$ includes the point $\displaystyle (-2,0)$, as can be seen in the diagram below:

Therefore, $\displaystyle f(-2) = 0$ and $\displaystyle g(0) =- f^{-1}(0)=- \left (-2 \right )= 2$. The $\displaystyle y$-intercept of the graph of $\displaystyle g$ is $\displaystyle (0, 2)$.

$\displaystyle |70-t| - 70 = 70$

$\displaystyle |70-t| - 70 + 70 = 70 + 70$

$\displaystyle |70-t| = 140$

Either 

$\displaystyle 70- t = 140$ or $\displaystyle 70- t = -140$

so we solve each separately.

$\displaystyle 70- t = 140$

$\displaystyle 70- t + t - 140 = 140 + t - 140$

$\displaystyle -70 = t$

$\displaystyle 70- t+ t + 140 = -140+ t + 140$

$\displaystyle 210 = t$

The solution set is $\displaystyle \left \{ -70, 210\right \}$.