Part (a) looks to maximize positive components in the solution vector x by solving a related LP. Part (b) looks to do the same with only 1 LP.
I am familiar with solving LPs, but I am not sure how to show that this problem is giving the maximum positive components.
For (a):
Suppose $x$ is an optimal solution. Then $cx$ with $c>0$ is also optimal. Now use $c$ to scale $x$ such that the smallest nonnegative element is at least $1$. Set $y_i = 1$ if $x_i>0$, $y_i=0$ otherwise. Let $z=x-y$ and note that $z$ is feasible.
Conversely, given an optimal solution $(y^*,z^*)$ of the LP, let $x=z^*+y^*$ and assume $x$ is not optimal. Then there exists a $v \neq 0$ such that $v+x \geq 0$ and $A(x+v)=0$. Now use $c>0$ to scale $x+v$ such that the smallest nonzero element is at least $1$ and construct $y$ as before. The sum of the elements of $y$ is the number of nonzero elements in $x+v$, while the sum of the elements of $y^*$ is the number of nonzero elements in $x$. The former being larger is a contradiction with optimality of $y^*$.
For (b) you change the right hand side from $0$ to $b$, and you modify $z\geq 0$ to $y+z\geq 0$. Then $z$ will be a solution to $Ax=b$, while $y$ still satisfies $Ay=0$.