(1) From the properties of an ellipse we know
$$B_1C_0 + C_0B_0 = B_1C_1 + C_1B_0.$$
Also, it is obvious that
$$B_0P_0 = B_0Q_0, \quad C_1B_0 + B_0Q_0 = C_1P_1,$$
$$B_1C_1 + C_1P_1 = B_1C_0 + C_0Q_1, \quad C_0Q_1 = C_0B_0 + B_0P'_0.$$
Adding these equations, we get $B_0P_0 = B_0P'_0$.
Therefore $P'_0$ and $P_0$ are coincident. Furthermore, as $P_0$, $C_0$ (the center of $\overarc{Q_1P_0}$) and $B_0$ (the center of $\overarc{P_0Q_0}$) are lying on the same line, we know that $\overarc{Q_1P_0}$ and $\overarc{P_0Q_0}$ are tangent at $P_0$.

(2) We have thus $\overarc{Q_1P_0}$ and $\overarc{P_0Q_0}$, $\overarc{P_0Q_0}$ and $\overarc{Q_0P_1}$, $\overarc{Q_0P_1}$ and $\overarc{P_1Q_1}$, $\overarc{P_1Q_1}$ and $\overarc{Q_1P'_0}$ are tangent at points $P_0, Q_0, P_1, Q_1$ respectively. Now we draw common tangent lines $P_0T$ and $P_1T$ through $P_0$ and $P_1$ respectively, and suppose the two lines meet at point $T$. Also, we draw a common

tangent line $R_1S_1$ through $Q_1$, and suppose it intercepts $P_0T$ and $P_1T$ at point $R_1$ and $S_1$ respectively. Drawing segments $P_0Q_1$ and $P_1Q_1$, we get isosceles triangles $P_0Q_1R_1$ and $P_1Q_1S_1$ respectively. Then we have
$$\begin{aligned} \angle P_0Q_1P_1 &= \pi - \angle P_0Q_1R_1 - \angle P_1Q_1S_1 \\ &= \pi - (\angle P_1P_0T - \angle Q_1P_0P_1) \\ &\quad - (\angle P_0P_1T - \angle Q_1P_1P_0). \end{aligned}$$
Since
$$\pi - \angle P_0Q_1P_1 = \angle Q_1P_0P_1 + \angle Q_1P_1P_0,$$
we obtain
$$\angle P_0Q_1P_1 = \pi - \frac{1}{2}(\angle P_1P_0T + \angle P_0P_1T).$$
In the same way, we can prove that
$$\angle P_0Q_0P_1 = \pi - \frac{1}{2}(\angle P_1P_0T + \angle P_0P_1T).$$
It implies that points $P_0, Q_0, Q_1, P_1$ are concyclic.