Solution

For equation $(i)$,

$wx+w(\overline{x}+y)+x(\overline{x}+y)$

$=wx+w\overline{x}+wy+x\overline{x}+xy$

Since,

$x\overline{x}=0$

So,

$=wx+w\overline{x}+wy+xy$

$=w(x+\overline{x})+wy+xy$

$=w+wy+xy$

$=w+xy$

But the right side is

$x+wy$

Since $w+xy$ is not always equal to $x+wy$, equation $(i)$ is false.

Now, for equation $(ii)$,

$(w\overline{x}(y+x\overline{z})+\overline{w}\overline{x})y$

$=(w\overline{x}y+w\overline{x}x\overline{z}+\overline{w}\overline{x})y$

Since,

$\overline{x}x=0$

So,

$=(w\overline{x}y+\overline{w}\overline{x})y$

$=w\overline{x}y+\overline{w}\overline{x}y$

$=\overline{x}y(w+\overline{w})$

$=\overline{x}y$

So, equation $(ii)$ is true.

Therefore, $(i)$ is false and $(ii)$ is true.