Let $\alpha,\beta$ be linear operators on a finite dimensional vector space $V$ over field $F$. Let $\gamma=\alpha\circ\beta$ and $\delta=\beta\circ\alpha$. Prove that:
(1). $m_\delta(x)$ divides $xM_\gamma(x)$. Here $m_T(x)$ denotes the minimal polynomial of the linear operator $T$.
(2). If $\lambda\neq 0$, then $Dim(Ker(\gamma-\lambda \cdot Id))=Dim(Ker(\delta-\lambda \cdot Id))$.
How to prove? I totally have no idea...
[I guess that $M_\gamma (x)$ actually means $m_\gamma (x)$, doesn't it?]
As for your first question, we could start with the following
Lemma 1. $(\beta\circ\alpha)\circ(\beta\circ\alpha)^n = \beta\circ(\alpha\circ\beta)^n\circ\alpha$.
Proof. Exercise. [Hint. Verify the statement for $n=1$ and then induction on $n$. Notice that it's also true for $n=0$ :-D ]
Thanks to this lemma, we are able to prove
Lemma 2. $xm_\gamma (x)$ annihilates $\delta$.
From which statement (1) follows because the minimal polynomial of $\delta$ divides every polynomial that annihilates $\delta$.
Proof of lemma 2. Let $m_\gamma (x) = x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$. Then if we evaluate $xm_\gamma (x)$ on $\delta$, we get
\begin{eqnarray*} \delta \circ m_\gamma (\delta ) &= \beta\circ\alpha \circ \left((\beta\circ\alpha)^n + a_{n-1} (\beta\circ\alpha)^{n-1}+ \dots + a_1(\beta\circ\alpha) + a_0\mathrm{Id} \right) \\ &= \beta \circ \left((\alpha\circ\beta)^n + a_{n-1} (\alpha\circ\beta)^{n-1}+ \dots + a_1(\alpha\circ\beta) + a_0\mathrm{Id} \right)\circ\alpha \end{eqnarray*}
The last equality follows from lemma 1. But now, inside the parenthesis, we have $m_\gamma(\gamma)$, which is zero. QED
As for your second question, I make the following claim:
Proposition. Assume $\lambda \neq 0$ is an eigenvalue of $\gamma$. Then, restricted to $\mathrm{ker}\ (\gamma - \lambda\mathrm{Id})$ and $\mathrm{ker}\ (\delta -\lambda\mathrm{Id} )$, we have $\alpha\circ\beta = \lambda\mathrm{Id}$ and $\beta\circ\alpha = \lambda\mathrm{Id}$.
Hence, restricted to those subspaces, $\alpha$ and $\beta$ are isomorphisms. In particular, their dimensions are equal.
Proof of the proposition. Exercise. [Hint: verify that, if $v$ is an eigenvector of $\delta$ with eigenvalue $\lambda$, then so is $\alpha (v)$ for $\gamma$. Analogously, if $w$ is an eigenvector of $\gamma$ with eigenvalue $\lambda$, then so is $\beta (w)$ for $\delta$. Then, compute the compositions $\alpha \circ \beta$ and $\beta \circ\alpha$ restricted to those subspaces.]