If $f\left(a\right)=\log \frac{2+a}{2-a}$, then $\frac{1}{2}f\left(\frac{8a}{4+a^2}\right)=$

$f\left(a\right)=\log \frac{2+a}{2-a}$

$\Rightarrow \frac{1}{2}f\left(\frac{8a}{4+a^2}\right)=\frac{1}{2}\log \left(\frac{{\displaystyle 2+\frac{{\displaystyle 8a}}{{\displaystyle 4+a^2}}}}{{\displaystyle 2-\frac{{\displaystyle 8a}}{{\displaystyle 4+a^2}}}}\right)$

$=\frac{1}{2}\log \left(\frac{8+2a^2+8a}{8+2a^2-8a}\right)$

$\Rightarrow \frac{1}{2}\log \left(\frac{a^2+4a+4}{a^2-4a+4}\right)=\frac{1}{2}\log {\left(\frac{2+a}{2-a}\right)}^2$

$=\log \left(\frac{2+a}{2-a}\right)$

$\therefore \frac{1}{2}f\left(\frac{8a}{4+a^2}\right)=f\left(a\right)$