By the row space method, the nonzero rows in reduced row echelon form a basis of the row space of $A$. Thus
From part (a), we see that the nullity of $A$ is $1$. The rank-nullity theorem says that
Since $A$ is in echelon form and it has two nonzero rows, the rank is $2$.
First of all, note that $A$ is already in reduced row echelon form.
Since the length of the basis vector is $sqrt{(-1)^2+0^2+1^2}=sqrt{2}$, it is not orthonormal basis.