Answer
a) $x^{n/2}$
b) $x^{(n-1)/2}*\sqrt x$
Work Step by Step
a) $n=2*y$ where $y$ a real number, so $n$ is even. By this logic, $n/2 = y$.
$\sqrt {x^n}$
$\sqrt {x^{2*y}}$
$x^y$
$x^{n/2}$
b)
$n=2*y+1$ where $y$ a real number, so $2y$ is even. $2y+1$ is odd, then. By this logic, $(n-1)/2=y$
$\sqrt {x^n}$
$\sqrt {x^{2*y}*x^1}$
$x^y*\sqrt x$
$x^{(n-1)/2}*\sqrt x$