1) Show that a linear functional is non-zero iff it is onto, and in this last case we always have that its kernel is a maximal proper subspace of $\,V\,$
2) So
$\ker T=\ker L\Longrightarrow \,\exists v\in V\,\,\,s.t.\,\,\,V=\langle\,\ker T\,,\,v\,\rangle=\langle\,\ker L\,,\,v\,\rangle\,\,,\,\,v\notin\ker T=\ker L$
Suppose $\,Tv=k\,\,,\,\,Lv=r$ and let $\,c\,$ be a solution to $\,xr=k\,$ ,thus
$Tv=k=cr=cLv=L(cv)$
and since obviously $\,T(u)=cL(u)=L(cu)=0\,\,\,,\,\,\forall\,u\in\ker T=\ker L\,$ , we are thus done.