Solution:

$y(n)-\frac{1}{4} y(n-1)-\frac{1}{8} y(n-2)=x(n)+3 x(n-1)+2 x(n-2)\\ $

Taking Z-transform,

$Y(Z)-\frac{1}{4} Z^{-1} Y(Z)-\frac{1}{8} Z^{-2} Y(Z)\\ $

$=X(Z)+3 Z^{-1} X(Z)+2 Z^{-1} X(Z)\\ $

$\frac{Y(Z)}{X(Z)}=\frac{1+3 Z^{-1}+2 Z^{-1}}{1-\frac{1}{4} Z^{-1}-\frac{1}{8} Z^{-2}}\\ $

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

$=H_1(Z) \cdot H_2(Z)\\ $

$H_1(Z)=\frac{\left(1+Z^{-1}\right)}{\left(1-\frac{1}{2} Z^{-1}\right)}\\ $

$ H_2(Z)=\frac{\left(1+2 Z^{-1}\right)}{\left(1+\frac{1}{4} Z^{-1}\right)}\\ …