$\left|\frac{z-4}{z-8}\right|=1$ and $\left|\frac{z}{z-2}\right|=\frac{3}{2}$
$\Rightarrow|z-4|=|z-8|$
Let $z=x+i y$ $|x+i y-4|=|x+i y-8|$
Squaring both sides, we get
$\left[(x-4)^{2}+y^{2}\right]=\left[(x-8)^{2}+y^{2}\right]$
$(x-4)^{2}=(x-8)^{2}$
$\Rightarrow x^{2}+16-8 x=x^{2}+64-16 x$
$\Rightarrow 8 x=48 \Rightarrow x=6$
when $\left|\frac{z}{z-2}\right|=\frac{3}{2}$
$\Rightarrow 2|z|=3|z-2|$
Squaring both sides, we get $4\left(x^{2}+y^{2}\right)=9\left[(x-2)^{2}+y^{2}\right]$
$\Rightarrow 4 x^{2}+4 y^{2}=9 x^{2}+36-36 x+9 y^{2}$
$\Rightarrow 5 x^{2}+5 y^{2}-36 x+36=0$
as we know $x=6$
$5(6)^{2}+5 y^{2}-36 \times 6+36=0$
$\Rightarrow 5 y^{2}=0 \Rightarrow y=0$
Hence $x=6$ and $y=0$. $\Rightarrow z=6$
$|z|=6$