$1\,)\,wx+w(x+y)+x(x+y)$
$=wx+wx+wy+x+xy$
$=wx+wy+x(1+y)=wx+wy+x$
$=x(1+w)+wy=x+wy$
So, $1$ is true.
$2\,)\,(w\bar{x}(y+x\bar{z})+\bar{w}\,\bar{x})y$
$=(w\bar{x}y+wx\bar{x}\bar{z}+\bar{w}\bar{x})y$
$=(\bar{x}(\bar{w}+wy)+0)y\,\,\,\left \{A\,\bar{A}=0\right \}$
$=(\bar{x}(\bar{w}+y))y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left \{A+\bar{A}B=A+B\right \}$
$=\bar{x}\bar{w}y+\bar{x}y=\bar{x}y(\bar{w}+1)$
$=\bar{x}y$
So, $2$ is true
$C$ should be the answer