$$\begin{gathered} Iff:[-6,6]\to Risdefinedbyf\left(x\right)=x^2-3\forall x\in R, \\ then\left(fofof\right)(-1)+\left(fofof\right)\left(0\right)+\left(fofof\right)\left(1\right)= \end{gathered}$$

$\begin{array}{l}f\left(x\right)=x^2-3\\ fofof(-1)=f\left(f\left(f(-1)\right)\right), fofof\left(0\right)=f\left(f\left(f\left(0\right)\right)\right), fofof\left(1\right)=f\left(f\left(f\left(1\right)\right)\right)\\ =f\left(f(-2)\right), =f\left(f(-3)\right), =f\left(f(-2)\right)\\ =f\left(1\right), =f\left(6\right), =f\left(1\right)\\ =-2, =33, =-2\\ \therefore fofof(-1)+fofof\left(0\right)+fofof\left(1\right)=-2+33-2\Rightarrow 29\\ 1)f\left(4\sqrt{2}\right)=32-3=29\end{array}$