Existence of a subspace with a certain property 
I am having trouble solving this problem.I have started solving the problem , so far my guesses for the subspace U were the intersection of V and complement of KerT , but i was soon able to come up with a counter example.
Then i made some modifications for U. My current guess for U is the following and i have proved that the intersection of U and Ker T is {0}. 

I am not confident that my guess for U is correct. 
 A: Notice first that $V \cap (V + (\ker T \cap \{0\}^c))^c=(V+(\ker T \cap \{0\}^c))^c$. This is because all sets are already contained in $V$, so the intersection is superfluous. For your proposed $U$, note that $V+(\ker T \cap \{0\}^c)=V$. Indeed, Given $x \in V$, choose $y \neq 0$ in $\ker T$. Then $-y \in V$. Hence $(x-y)+y=x \in V+ (\ker T \cap \{0\}^c).$ This shows that $(V+(\ker T \cap \{0\})^c)^c = \emptyset$.
In general, it is difficult to build subspaces by taking set-theoretic complements and removing points. A good way to get a concrete handle on problems like this is to work with a basis for your vector space. By choosing a convenient basis, this particular problem simplifies considerably.
Let $\{e_1,\dots,e_m\}$ be a basis for $\ker T$; then you can extend this set to a basis $\{e_1,\dots,e_k\}$ for $V$. Assume $\ker T \neq V$ (this case is easily dealt with separately), so that $k>m$. Let $$U=\langle e_{m+1},\dots,e_k \rangle.$$ 
From the fact that $e_i$'s form a basis, you can deduce that $U \cap \ker T= \{0\}$ and that $T(U)=\operatorname{Range}T.$ For proving the part about the image, the rank nullity theorem might also come in handy.
