With the incomplete base theorem : take $(v_1,...,v_p)$ a base of $\ker T\subset V.$ Now complete it in a base $(v_1,...,v_p,v_{p+1},...,v_n)$ of $V.$ You will have that $$T(V)=\underset{=\{0\}}{\underbrace{T(v_1)\oplus T(v_2)\oplus...\oplus T(v_p)}}\oplus T(v_{p+1})\oplus...\oplus T(v_n)=T(v_{p+1})\oplus...\oplus T(v_n)$$ and as $T$ is injective on $\mathrm{span}\,(v_{p+1},...,v_n)$ you get that $(T(v_{p+1}),...,T(v_n))$ is a base of $T(V).$ Finally, $$\dim\ker T+\mathrm{rank}\,T=p+(n-(p+1)+1)=n=\dim V.$$
To get you started, show that
$$\operatorname{ran}(L^k) = \operatorname{ran} (L^{k+1}) = \operatorname{ran} (L^{k + 2}) = \ldots$$
In particular, try using the fact that $\operatorname{ran} (L^k) = \operatorname{ran} (L^{2k})$. This means that, given any $v \in V$, there exists a $w \in V$ such that $L^kv = L^{2k}w$. Try using $w$ to find a decomposition of $v$ into a sum from the kernel and range of $L^k$.
Then, try showing that the intersection of the kernel and range is trivial. For this, you'll want to use the fact that $\operatorname{ker} (L^k) = \operatorname{ker} (L^{2k})$.
EDIT:
Because $\operatorname{rank}(L^k) = \operatorname{rank} (L^{k+1})$ and $\operatorname{ran}(L^k) \supseteq \operatorname{ran} (L^{k+1})$, we must have $\operatorname{ran}(L^k) = \operatorname{ran} (L^{k+1})$. By the rank-nullity theorem, we also have $\operatorname{null}(L^k) = \operatorname{null} (L^{k+1})$. Since we have $\operatorname{ker}(L^k) \subseteq \operatorname{ker} (L^{k+1})$, we similarly get $\operatorname{ker}(L^k) = \operatorname{ker} (L^{k+1})$.
Moreover, this equality continues to hold as we increase $k$. This is a standard result in linear algebra. Again, we have that $\operatorname{ran}(L^{k+1}) \supseteq \operatorname{ran}(L^{k+2})$, so we just need to show that, given a vector $v \in \operatorname{ran}(L^{k+1})$, then $v \in \operatorname{ran}(L^{k+2})$.
If $v \in \operatorname{ran}(L^{k+1})$, then there exists some $w \in V$ such that $v = L^{k+1}w$. Note that
$$L^k w \in \operatorname{ran}(L^k) = \operatorname{ran}(L^{k+1}),$$
so there must exist some $u \in V$ such that $L^k w = L^{k+1} u$. Therefore,
$$v = L^{k + 1} w = L(L^k w) = L(L^{k + 1} u) = L^{k + 2} u \in \operatorname{ran}(L^{k+2}),$$
as required. Therefore, $\operatorname{rank}(L^{k+1}) = \operatorname{rank} (L^{k+2})$, and so, by the previous logic, we also get $\operatorname{ker}(L^{k+1}) = \operatorname{ker} (L^{k+2})$.
You can continue this by induction to show that
\begin{align*}
\operatorname{ran}(L^k) &= \operatorname{ran} (L^{k+1}) = \operatorname{ran} (L^{k+2}) = \ldots \\
\operatorname{ker}(L^k) &= \operatorname{ker} (L^{k+1}) = \operatorname{ker} (L^{k+2}) = \ldots
\end{align*}
In particular, as I said, we need the fact that $\operatorname{ran}(L^k) = \operatorname{ran} (L^{2k})$ and $\operatorname{ker}(L^{2k}) = \operatorname{ker} (L^{2k})$.
Now we show $V = \operatorname{ker}(L^k) \oplus \operatorname{ran}(L^k)$. Let $v \in V$. Note that
$$L^k v \in \operatorname{ran}(L^k) = \operatorname{ran}(L^{2k})$$
so there must exist some $w \in V$ such that $L^{2k}w = L^k v$. Then, we have
$$v = (v - L^k w) + L^k w,$$
where $L^k w \in \operatorname{ran}(L^k)$ and $v - L^k w \in \operatorname{ker}(L^k)$, as
$$L^k(v - L^k w) = L^k v - L^{2k} w = 0.$$
Therefore, $V = \operatorname{ker}(L^k) + \operatorname{ran}(L^k).$
To show the sum is direct, it suffices to show $\operatorname{ker}(L^k) \cap \operatorname{ran}(L^k) = \{ 0 \}$. Suppose $v \in \operatorname{ker}(L^k) \cap \operatorname{ran}(L^k)$. Then $v = L^k w$ for some $k$ and $0 = L^k v = L^{2k} w$. Therefore $w \in \operatorname{ker}(L^{2k}) = \operatorname{ker}(L^k)$, so $0 = L^k w = v$,
as required.
Best Answer
(a) Note that $\ker(T) \subset \ker(T^2)$ because $T(x) = 0$ implies $T(T(x)) = 0$. The equality you're positing is that $\dim \ker(T) = \dim \ker(T^2)$, by the rank-nullity theorem. It follows that $\ker(T) = \ker(T^2)$.
So now assume that $x \in \ker(T) \cap \operatorname{im}(T)$. Then $x = T(x')$ for some $x'$, because $x \in \operatorname{im}(T)$. But then $T(x) = T(T(x')) = 0$. It follows from $\ker(T) = \ker(T^2)$ that $T(x') = 0$, which is to say, $x = 0$.
For (b), see the following question.