Let  $x_{1}and{x}_2$ be solutions of the equation  ${\sin }^{-1}(x^2-3x+\frac{5}{2})=\frac{\pi }{6}$. Then, the value of  $x_1^2+x_2^2$  is

${\sin }^{-1}(x^2-3x+\frac{5}{2})=\frac{\pi }{6}$

$\Rightarrow x^2-3x+\frac{5}{2}=\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$

$\Rightarrow 2x^2-6x+5=1$

$\Rightarrow 2x^2-6x+4=0$

$x^2-3x+2=0$

Since,  $x_1,x_2$ are the roots

$\therefore x_1+x_2=3and{x}_1{x}_2=2$

$\therefore x_1^2+x_2^2={(x_1+x_2)}^2-2x_2{x}_2=9-2\left(2\right)=5$.