Distance between two parallel planes $2x + y + 2z = 8 $ and $4x + 2y + 4z + 5 = 0 $, is

$\frac{7}{2}$

We know that the distance between parallel planes $ax + by + cz + d_1 = 0 $ and $ax + by + cz + d_2 = 0 $ is given by 

$\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$

Here, the planes are $2x + y + 2z - 8 = 0 $ and $2x + y + 2z +\frac{5}{2} = 0 $

∴ Required distance $=\begin{vmatrix} \frac{\begin{vmatrix}-8-\frac{5}{2}\end{vmatrix}}{\sqrt{2^2+1^2+2^2}}\end{vmatrix}=\frac{7}{2}$