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