Let $p$ be an odd prime greater than 3. Find
(1) $\sum_{i=0}^{p} \mathrm{C}_{2 i}^{i}(\bmod p)$;
(2) $\sum_{i=0}^{2 p} \mathrm{C}_{2 i}^{i}(\bmod p)$.
【Analysis】(1) Note that,
$$
\sum_{i=0}^{p} \mathrm{C}_{2 i}^{i}=\sum_{i=0}^{\frac{p-1}{2}} \mathrm{C}_{2 i}^{i}+\sum_{i=\frac{p+1}{2}}^{p-1} \mathrm{C}_{2 i}^{i}+\mathrm{C}_{2 p}^{p} .
$$

By Example 2, the first term $\sum_{i=0}^{\frac{p-1}{2}} \mathrm{C}_{2 i}^{i}$, when divided by $p$, leaves a remainder of either 1 or -1.
Next, consider the second term $\sum_{i=\frac{p+1}{2}}^{p-1} \mathrm{C}_{2 i}^{i}$.
Since $\frac{p+1}{2} \leqslant i \leqslant p-1$, then $2 i \geqslant p+1$, and $p \mid (2 i)!$.
By $\mathrm{C}_{2 i}^{i}=\frac{(2 i)!}{i! \cdot i!}$, i.e., $(2 i)!=i! \cdot i! \mathrm{C}_{2 i}^{i}$, we know $p \mid i! \cdot i! \mathrm{C}_{2 i}^{i}$.
However, $p \nmid i!$, so $p \mid \mathrm{C}_{2 i}^{i}$.
Therefore, $\sum_{i=\frac{p+1}{2}}^{p-1} \mathrm{C}_{2 i}^{i} \equiv 0(\bmod p)$.
Next, analyze the third term $\mathrm{C}_{2 p}^{p}$.
Since $\mathrm{C}_{2 p}^{p}=2 \mathrm{C}_{2 p-1}^{p-1}$, we have,
$$
\begin{array}{l}
(p-1)!\mathrm{C}_{2 p}^{p}=2(2 p-1) \cdots(p+1) \\
\equiv 2(p-1)(p-2) \cdots 1 \\
\equiv 2((p-1)!)(\bmod p) .
\end{array}
$$

Then $p!(p-1)!\left(\mathrm{C}_{2 p}^{p}-2\right)$, and $p \nmid (p-1)!$.
Hence $p \mid \left(\mathrm{C}_{2 p}^{p}-2\right)$, i.e., $\mathrm{C}_{2 p}^{p}=2(\bmod p)$.
In summary, $\sum_{i=0}^{p} \mathrm{C}_{2 i}^{i}(\bmod p)=3$ or 1.
(2) Note that,
$$
\begin{array}{l}
\sum_{i=0}^{2 p} \mathrm{C}_{2 i}^{i}=\sum_{i=0}^{p} \mathrm{C}_{2 i}^{i}+\sum_{i=p+1}^{2 p} \mathrm{C}_{2 i}^{i} \\
=\sum_{i=0}^{p} \mathrm{C}_{2 i}^{i}+\sum_{i=1}^{p} \mathrm{C}_{2 p+2 i}^{p+i} .
\end{array}
$$

When $1 \leqslant i \leqslant \frac{p-1}{2}$, by Lucas's Theorem, $\mathrm{C}_{2 p+2 i}^{p+i} \equiv \mathrm{C}_{2}^{1} \mathrm{C}_{2 i}^{i} \equiv 2 \mathrm{C}_{2 i}^{i}(\bmod p)$ $\Rightarrow \sum_{i=1}^{\frac{p-1}{2}} \mathrm{C}_{2 p+2 i}^{p+i} \equiv 2 \sum_{i=1}^{\frac{p-1}{2}} \mathrm{C}_{2 i}^{i}(\bmod p)$.
The right side of the above equation, when divided by $p$, leaves a remainder of 2 or -2.
When $\frac{p-1}{2}<i \leqslant p-1$, $(p+i)+(p+i)$ must carry once in base $p$, by Kummer's Theorem,
$$
p \mid \mathrm{C}_{2 p+2 i}^{p+i}, \sum_{i=\frac{p+1}{2}}^{p-1} \mathrm{C}_{2 p+2 i}^{p+i} \equiv 0(\bmod p) .
$$

By Lucas's Theorem,
$$
\mathrm{C}_{4 p}^{2 p} \equiv \mathrm{C}_{4}^{2} \mathrm{C}_{0}^{0} \equiv 6(\bmod p) \text {. }
$$

And $\sum_{i=0}^{p} \mathrm{C}_{2 i}^{i}(\bmod p)=3$ or 1, then
$$
\sum_{i=0}^{2 p} \mathrm{C}_{2 i}^{i}(\bmod p)=11 \text { or } 5 \text {. }
$$