$f(x)=\frac{{{x}^{2}}}{8}+1\Rightarrow f(8)=8+1=9$  ${f}'(x)=\frac{x}{4}\Rightarrow {f}'(8)=2$  $g(x\sqrt{x})=\frac{{{x}^{2}}}{8}+1\xrightarrow{x=4}g(8)=\frac{16}{8}+1=3$  $(g(x\sqrt{x}){)}'=(\sqrt{x}+\frac{x}{2\sqrt{x}}){g}'(x\sqrt{x})=\frac{x}{4}\xrightarrow{x=4}(2+\frac{4}{4}){g}'(8)=1$  $3{g}'(8)=1\Rightarrow {g}'(8)=\frac{1}{3}$  $(fg{)}'(8)={f}'(8)g(8)={g}'(8)f(8)\Rightarrow (fg{)}'(8)=2\times 3+\frac{1}{3}\times 9=6+3=9$