How can we find z when the complex number Z satisfies the equation 4z3z―=(1−8i2−i),where z―...

The question would make more sense if it is about
${\displaystyle 4z+3\bar{z}=\frac{1-8i}{2-i}}$
With the product in the left-hand side the equation would obviously have no solution, because the left-hand side is real and the right-hand side isn’t.
Write the right-hand side in rectangular form:
${\displaystyle \frac{1-8i}{2-i}=\frac{(1-8i)(2+i)}{4+1}=\frac{2-16i+i-8i^2}{5}=2-3i}$
Then the equation implies, taking conjugates
$$\{\begin{array}{l}4z+3\bar{z}=2-3i\\ 4\bar{z}+3z=2+3i\end{array}$$
Summing up we obtain ${\displaystyle 7(z+\bar{z})=4}$. Subtracting we get ${\displaystyle z-\bar{z}=-6i}$. Therefore
${\displaystyle z+\bar{z}=\frac{4}{7},z-\bar{z}=-6i}$
and finally ${\displaystyle 2z=\frac{4}{7}-6i}$

Let $Z=x+iy$
$$\begin{gathered} \therefore 4(x+iy)3(x-iy)=\frac{1-8i}{2-i} \\ \Rightarrow 12(x^2+y^2)=\frac{(1-8i)(2+i)}{(2-i)(2+i)}=\frac{2+i-16i+8}{4+1} \\ \Rightarrow 12(x^2+y^2)=\frac{10-15i}{5}=2-3i \end{gathered}$$
Left hand side is real. It can not be equal to a complex number. Hence there is no z satisfying the given condition.