Answer
$(a)$
$x$-intercept, $A(8,0)$
$y$-intercept, $B(0,-2)$
$(b)$
$x$-intercepts, $A(-\sqrt{3},0)$; $B(\sqrt{3},0)$
$y$-intercept, $C(0,3)$
Also see the images below.
Work Step by Step
$(a)$ $x-4y=8$, We have a linear equation.
For $x$-intercept, $y=0$
$x-4\times0=8$
$x=8$
$A(8,0)$
For $y$-intercept, $x=0$
$0-4y=8$
$4y=-8$
$y=-2$
$B(0,-2)$
For a better visualization (while sketching the graph), let's find one more random point on the line:
$y=-1$
$x-4\times(-1)=8$
$x+4=8$
$x=4$
$C(4,-1)$
To sketch the graph, we simply plot these points and connect them. See the image above.
$(b)$ $y=-x^2+3$, In this case we have a quadratic equation.
For $x$-intercept, $y=0$
$0=-x^2+3$
$x^2=3$
$x=±\sqrt{3}$
$A(-\sqrt{3},0)$; $B(\sqrt{3},0)$
For $y$-intercept, $x=0$
$y=-0^2+3$
$y=3$
$C(0,3)$
Quadratic equation has a form of parabola. To graph the parabola we need some critical points. $x$-intercepts, $y$-intercept and vertex coordinates.
$V(-\frac{b}{2a}, -\frac{D}{4a})$
$x=-\frac{b}{2a}=-\frac{0}{2}=0$
$y=-\frac{D}{4a}=-\frac{-12}{4}=3$
$V(0,3)$
We can plot these points and sketch a parabola. See the image below
