6. 利用平方差公式简便计算：
(1) $107×93$；
(2) $2022^{2}-2021×2023$。

(1)
$\begin{aligned}107 × 93 &= (100 + 7)(100 - 7) \\&= 100^2 - 7^2 \\&= 10000 - 49 \\&= 9951\end{aligned}$
(2)
$\begin{aligned}2022^2 - 2021 × 2023 &= 2022^2 - (2022 - 1)(2022 + 1) \\&= 2022^2 - (2022^2 - 1^2) \\&= 2022^2 - 2022^2 + 1 \\&= 1\end{aligned}$

7. 阅读材料后解决问题：
计算$(2 + 1)(2^{2}+1)(2^{4}+1)(2^{8}+1)$。经过观察，小明发现如果将原式进行适当的变形后可以出现特殊的结构，进而可以应用平方差公式解决问题，具体解法如下：
$\begin{aligned}&(2 + 1)(2^{2}+1)(2^{4}+1)(2^{8}+1)\\=&(2 - 1)(2 + 1)(2^{2}+1)(2^{4}+1)(2^{8}+1)\\=&(2^{2}-1)(2^{2}+1)(2^{4}+1)(2^{8}+1)\\=&(2^{4}-1)(2^{4}+1)(2^{8}+1)\\=&(2^{8}-1)(2^{8}+1)\\=&2^{16}-1.\end{aligned}$
根据上述内容，试着解决以下的问题。
(1) $(2 + 1)(2^{2}+1)(2^{4}+1)(2^{8}+1)(2^{16}+1)=$；
(2) 计算$(3 + 1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)$的值。

(1)
$\begin{aligned}&(2 + 1)(2^{2}+1)(2^{4}+1)(2^{8}+1)(2^{16}+1)\\=&(2 - 1)(2 + 1)(2^{2}+1)(2^{4}+1)(2^{8}+1)(2^{16}+1)\\=&(2^{2}-1)(2^{2}+1)(2^{4}+1)(2^{8}+1)(2^{16}+1)\\=&(2^{4}-1)(2^{4}+1)(2^{8}+1)(2^{16}+1)\\=&(2^{8}-1)(2^{8}+1)(2^{16}+1)\\=&(2^{16}-1)(2^{16}+1)\\=&2^{32}-1\end{aligned}$
(2)
$\begin{aligned}&(3 + 1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1) \\ &=\frac{1}{2}(3 - 1)(3 + 1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)\\&=\frac{1}{2}(3^{2}-1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)\\&=\frac{1}{2}(3^{4}-1)(3^{4}+1)(3^{8}+1)(3^{16}+1)\\&=\frac{1}{2}(3^{8}-1)(3^{8}+1)(3^{16}+1)\\&=\frac{1}{2}(3^{16}-1)(3^{16}+1)\\&=\frac{3^{32}-1}{2}\end{aligned}$