Answer
$a_n=\sqrt2^n$ and $a_5=4\sqrt2.$
Work Step by Step
We know that if $a_1$ is the first term and $r$ is the common ratio, then the nth term is: $a_n=a_1\cdot r^{n-1}.$
Hence here $a_n=\sqrt2\cdot \sqrt2^{n-1}=\sqrt2^n$ and $a_{5}=\sqrt2^5=4\sqrt2.$