If  $f^{\text{'}\text{'}}\left(x\right)=-f\left(x\right)$ and  $g\left(x\right)=f^{\text{'}}\left(x\right)$  and  $F\left(x\right)={\left(f\left(\frac{x}{2}\right)\right)}^2+{\left(g\left(\frac{x}{2}\right)\right)}^2$ and given that  $F\left(5\right)=5$, then  $F\left(10\right)$ is equal to

$f^{\text{'}\text{'}}\left(x\right)=-f\left(x\right) g\left(x\right)=f^{\text{'}}\left(x\right)$

$f^{\text{'}}\left(x\right)=2f\left(\frac{x}{2}\right)+f^{\text{'}}\left(\frac{x}{2}\right)\frac{1}{2}+2g\left(\frac{x}{2}\right)g^{\text{'}}\left(\frac{x}{2}\right)\frac{1}{2}=f\left(\frac{x}{2}\right)f^{\text{'}}\left(\frac{x}{2}\right)+f^{\text{'}}\left(\frac{x}{2}\right)(-f\left(\frac{x}{2}\right))=0$

 $f\left(x\right)=$ constant function
 $f\left(x\right)=5$

$$\begin{gathered} f\left(5\right)=\text{constant}=5 \\ So, f(x=10)=5 \end{gathered}$$