jmc

We begin by finding $f\left(\frac{\pi }{3}\right)$. Since we know that $\pi \approx 3.14$, $\frac{\pi }{3}$ must be slightly greater than $1$, so $f\left(\frac{\pi }{3}\right)=\lceil \frac{\pi }{3}\rceil =2$. To find $f(\sqrt{45})$, we realize that $\sqrt{36}<\sqrt{45}<\sqrt{49}$, so $6<\sqrt{45}<7$. Therefore, since $\sqrt{45}\ge 4$, we have that $f(\sqrt{45})=\lfloor \sqrt{45}\rfloor =6$. Finally, we consider $f(8^{2/3})$. We know that $8^{2/3}=(\sqrt[3]{8}{)}^2=2^2=4$, so $f(8^{2/3})=\lfloor 8^{2/3}\rfloor =4$. Thus, we have that $f\left(\frac{\pi }{3}\right)+f(\sqrt{45})+f(8^{2/3})=2+6+4=\boxed{12}$.