#### Answer

See the explanation below.

#### Work Step by Step

a) To find the $y$ interpret , we put $x=0$
Thus, $\dfrac{y^2}{9}-\dfrac{0^2}{16}=1$
or, $y^2 =9$
or, $y = \pm 3$
Therefore, $y$ intercepts are: {$-3,3$}
b)To find the $x$ interpret , we put $y=0$
Thus, $\dfrac{0^2}{9}-\dfrac{x^2}{16}=1$
or, $x^2 = -16$
or, $x = \pm \sqrt{-16}$
Therefore, there are no $x$ intercepts because of of no real numbers or a non-real solution.