jmc

Let $z=x+yi,$ where $x$ and $y$ are real numbers.

From the equation $\left|\frac{z-4}{z-8}\right|=1,$ $|z-4|=|z-8|.$ Then $$|x+yi-4|=|x+yi-8|,$$so $(x-4{)}^2+y^2=(x-8{)}^2+y^2.$ This simplifies to $x=6.$

From the equation $\left|\frac{z-12}{z-8i}\right|=\frac{5}{3},$ $3|z-12|=5|z-8i|.$ Then $$3|6+yi-12|=5|6+yi-8i|,$$so $9(36+y^2)=25(36+(y-8{)}^2).$ This simplifies to $16y^2-400y+2176=0,$ which factors as $16(y-8)(y-17)=0.$ Hence, $y=8$ or $y=17.$

Thus, the solutions in $z$ are $\boxed{6+8i,6+17i}.$