If you have orthonormal vectors $q_1,q_2,q_3,q_4$ of $\mathbb R^4$ and orthonormal vectors $z_1,z_2,z_3,z_4,z_5,z_6$ of $\mathbb R^6$ and matrix A is $A=z_1q_1^T+z_2q_2^T$.
a)Find a base and dimension of fundamental subspaces of $A$
b) Using fact from a) find a solution $Ax=z_1$
For a)$ Im(A)=L(z_1,z_2), dimIm(A)=2, Im(A^T)=L(q_1,q_2), dim Im(A^T)=2, ker(A)=L(q_3,q_4) dimker(A)=2, ker(A^T)=L(z_3,z_4,z_5,z_6) ,dim N(A^T)=4.$
For b) $Ax=z_1$, If $z_1 \in Im(A)$ then this equation have solution, then $x=q_1$ since $(x_1,x_1)=1$ and $(x_2,x_1)=0$ so $Aq_1=(z_1q_1^Tq_1+z_2q_2^Tq_1)=z_1+0=z_1$ is this ok?
It looks like you’ve gotten the domain and codomain swapped. We have $$Av = z_1q_1^Tv+z_2q_2^Tv = (q_1^Tv)\,z_1+(q_2^Tv)\,z_2$$ since the parenthesized quantities are scalars. For those products to make sense we must have $v\in\mathbb R^4$. At the same time, we can see from the above that the result of multiplying any vector $v$ by $A$ is a linear combination of $z_1$ and $z_2$, which live in $\mathbb R^6$. I expect that you’ll be able to take it from here.