Use L'Hopital's rule twice:
[tex]\displaystyle \lim_{x\to0^+} \frac1{\sin^2(x)} \int_{\frac x2}^x \sin^{-1}(t) \, dt = \lim_{x\to0^+} \frac1{\sin^2(x)} \left(\int_0^x \sin^{-1}(t) \, dt - \int_0^{\frac x2} \sin^{-1}(t) \, dt\right) \\\\ = \lim_{x\to0^+} \frac{\sin^{-1}(x) - \frac12 \sin^{-1}\left(\frac x2\right)}{2\sin(x)\cos(x)} ~~~~~~~~~~ (LHR) \\\\ = \lim_{x\to0^+} \frac{2\sin^{-1}(x) - \sin^{-1}\left(\frac x2\right)}{2\sin(2x)} \\\\ = \lim_{x\to0^+} \frac{\frac2{\sqrt{1-x^2}} - \frac1{2\sqrt{1-\frac{x^2}4}}}{4\cos(2x)} ~~~~~~~~~~ (LHR) \\\\ = \lim_{x\to0^+} \frac{\frac2{\sqrt{1-x^2}} - \frac1{\sqrt{4-x^2}}}{4\cos(2x)}[/tex]
The remaining limand is continuous at x = 0, and the limit is
[tex]\displaystyle \lim_{x\to0^+} \frac1{\sin^2(x)} \int_{\frac x2}^x \sin^{-1}(t) \, dt = \frac{2-\frac12}4 = \boxed{\frac38}[/tex]
