Answer
$a)$ Domain and range of this relation is all real numbers.
$b)$
$$f(x)=\frac14x+\frac32$$
$$f(-2)=1$$
Work Step by Step
$$x-4y=-6$$
Isolate $y$:
$$-4y=-x-6$$
$$y=\frac14x+\frac32$$
This equation is a linear and represents line.
$a)$ Since it is a line, domain and range of this relation is all real numbers.
$b)$ Since it is a line this relation is indeed a function.
Rewrite as:
$$f(x)=\frac14x+\frac32$$
Substitute $x=-2$ into the function:
$$f(-2)=\frac14(-2)+\frac32$$
Compute the right side:
$$f(-2)=1$$
