#### Answer

$a_5=-32 \text{ and } a_n=-\dfrac{1}{8}\cdot4^{n-1}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To find $a_5$ and $a_n$ for the given geometric sequence described as \begin{array}{l}\require{cancel} a_3=-2 \\ r=4 ,\end{array} use the formula for finding the $n^{th}$ term of a geometric sequence.
$\bf{\text{Solution Details:}}$
Using the formula for finding the $n^{th}$ term of a geometric sequence, which is given by $a_n=a_1r^{n-1},$ then \begin{array}{l}\require{cancel} a_n=a_1r^{n-1} \\\\ a_3=a_1r^{3-1} \\\\ a_3=a_1r^{2} .\end{array}
Subtituting the given values, $a_3=-2$ and $r=4,$ the equation above results to \begin{array}{l}\require{cancel} -2=a_1(4^{2}) \\\\ -2=a_1(16) \\\\ -\dfrac{2}{16}=a_1 \\\\ a_1=-\dfrac{1}{8} .\end{array}
With $a_1=-\dfrac{1}{8}$ and $r=4,$ then \begin{array}{l}\require{cancel} a_n=a_1r^{n-1} \\\\
a_n=-\dfrac{1}{8}\cdot4^{n-1}
.\end{array}
With $n=5,$ the equation above bcomes \begin{array}{l}\require{cancel}
a_5=-\dfrac{1}{8}\cdot4^{5-1}
\\\\
a_5=-\dfrac{1}{8}\cdot4^{4}
\\\\
a_5=-\dfrac{1}{8}\cdot256
\\\\
a_5=-1(32)
\\\\
a_5=-32
.\end{array}
Hence, $
a_5=-32 \text{ and } a_n=-\dfrac{1}{8}\cdot4^{n-1}
.$