#### Answer

$\:(-2,0)\cup(4,\infty)$

#### Work Step by Step

$\mathrm{See\:the\:graph\:above.}$
$a.\:\:\:$ By observing the graph, we come to know that the point of intersections are $\:(-2,-1)\:$ and $\:(4,2).$ Based on the graph, the function $\:y=\frac{x}{2}\:$ (represented by thick black line) is above the function $\:y=1+\frac{4}{x}\:$ (represented by red thick curve), when:
$-2 < x < 0\:\:$ or $\:\:x>4$
$b.\:\:\:$ $\frac{x}{2}>1+\frac{4}{x}$
$\frac{x}{2}-\frac{4}{x}-1>0$
$\frac{x^2-8-2x}{2x}>0$
Fraction is positive if both numerator and denominator are positive.
$\mathrm{Case\:I:}$
$x^2-8-2x > 0,\:$ zeros of the function are -2 and 4, and the leading coefficient is positive. So, we got a parabola opening upwards, meaning that it is $\:> 0\:$ on $(\:-\infty,-2)\:$ and $(\:4,\infty).$
$2x> 0\:$ when $\:a>0$
We need both to be satisfied, so in the first case we have $\:x\in(4,\infty).$
$\mathrm{Case\:II:}$
$x^2-8-2x < 0\:$ when $(-2,4)$
$2x < 0\:$ when $\:x\in(-\infty,0)$
Solutions of this case are on $\:(-2,0)$
Combined solution is $\:(-2,0)\cup(4,\infty)$