Endomorphisms: if $Im(g)$ is contained in $Im(f)$ then necessarily $g=f \phi$? Let $f$ and $g$ be endomorphisms of a vector space $V$. 
If $Im(g)$ is contained in $Im(f)$ then does there necessarily exist an endomorphism $\phi$ of $V$ such that  $g=f \phi$?
Do we also have that $Ker(g)$ contained in $Ker(f)$ implies the existence of an endomorphism $\phi$ such that $f=\phi g$?

Hint : The hypothesis already tells you that for all $v\in V$, $g(v)\in Im(f)$, i.e. there exist $w\in V$ such that $f(w)=g(v)$. Use this to construct first a function $\phi$ on a basis of $V$, and then extend that function to a linear application defined on $V$.
For the second part, first take a basis $\mathcal{B}_1$ of $\ker g$, and then extend it to a basis $\mathcal{B}$ of $V$. Then the complement $\mathcal{B}_2=\mathcal{B}\setminus \mathcal{B}_1$ is a linearly independent family, and $g(\mathcal{B}_2 )$ is a basis of the image of $g$. Extend it to a basis $\mathcal{B}'$ of $V$, and then you can simply define $\phi $ by putting $\phi(g(b))=f(b)$ for all $b\in \mathcal{B}_2$ and anything you want for the other vectors in your basis $\mathcal{B}'$. You can now check that $f=\phi\circ g$ by comparing the values on $\mathcal{B}$ : on $\mathcal{B}_1$ both are zero by hypothesis, on $\mathcal{B}_2$ they are equal by construction.
