[tex]x[/tex] is in quadrant I, so [tex]0<x<\dfrac\pi2[/tex], which means [tex]0<\dfrac x2<\dfrac\pi4[/tex], so [tex]\dfrac x2[/tex] belongs to the same quadrant.
Now,
[tex]\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}[/tex]
Since [tex]\sin x=\dfrac5{13}[/tex], it follows that
[tex]\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}[/tex]
Since [tex]x[/tex] belongs to the first quadrant, you take the positive root ([tex]\cos x>0[/tex] for [tex]x[/tex] in quadrant I). Then
[tex]\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}[/tex]
[tex]\tan x[/tex] is also positive for [tex]x[/tex] in quadrant I, so you take the positive root again. You're left with
[tex]\tan\dfrac x2=\dfrac15[/tex]
