To check whether each point is a solution to the equation, we substitute it into the equation and see if it results in a true statement.

First, we substitute $(-8,-2)$ into the equation:

\begin{align*} y &= \dfrac{x}{4} \\ (-2) &= \dfrac{-8}{4}\\ -2 &= -2\; {\color{green}{\checkmark}} \end{align*}

The result is a true statement, so $(-8,-2)$ is a solution to the equation.

Next, we substitute $(4,1)$ into the equation:

\begin{align*} y &= \dfrac{x}{4} \\ (1) &= \dfrac{4}{4}\\ 1 &= 1\; {\color{green}{\checkmark}} \end{align*}

The result is again a true statement, so $(4,1)$ is also a solution to the equation.

Finally, we substitute $(-4,1)$ into the equation:

\begin{align*} y &= \dfrac{x}{4} \\ (1) &= \dfrac{-4}{4} \\ 1 &= -1\; {\color{red}{\times}} \end{align*}

The result is a false statement, so $(-4,1)$ is not a solution to the equation.

So, out of the three ordered pairs given, the only pair that is not a solution to the equation is III $(-4,1).$

To check whether each point is a solution to the equation, we substitute it into the equation and see if it results in a true statement.

First, we substitute $(8,12)$ into the equation:

\begin{align*} y &= x+4 \\ (12) &= 8+4 \\ 12 &= 12 \; {\color{green}{\checkmark}} \end{align*}

The result is a true statement, so $(8,12)$ is a solution to the equation.

Next, we substitute $(-4,-8)$ into the equation:

\begin{align*} y &= x+4 \\ (-8) &= -4+4 \\ -8 &= 0 \; {\color{red}{\times}} \end{align*}

The result is a false statement, so $(-4,-8)$ is not a solution to the equation.

Finally, we substitute $(-6,-2)$ into the equation:

\begin{align*} y &= x+4 \\ (-2) &= -6+4 \\ -2 &= -2 \; {\color{green}{\checkmark}} \end{align*}

The result is a true statement, so $(-6,-2)$ is a solution to the equation.

So, out of the three ordered pairs given, the only pairs that are solutions to the equation are I $(8,12)$ and III $(-6,-2).$

Since $(4,t)$ is a solution to the linear equation $y=3x-2,$ the equation must turn into a true statement if we substitute $4$ and $t$ for $x$ and $y.$ So, we replace $x$ with $4$ and $y$ with $t,$ as follows:

\begin{align*} y &= 3x-2 \\ (t) &= 3(4)-2 \\ t &= 12-2 \\ t &= 10 \end{align*}

Since $(6,t)$ is a solution to the linear equation $y=-3x-7,$ the equation must turn into a true statement if we substitute $6$ and $t$ for $x$ and $y.$ So, we replace $x$ with $6$ and $y$ with $t,$ as follows:

\begin{align*} y &= -3x-7 \\ (t) &= -3(6)-7 \\ t &= -18-7 \\ t &= -25 \end{align*}