a) Đặt \(\left(a+b,a-b\right)=d\)

\(\Rightarrow\left\{{}\begin{matrix}a+b⋮d\\a-b⋮d\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a⋮d\\2b⋮d\end{matrix}\right.\). Do \(\left(a,b\right)=1\) nên từ đây suy ra \(d\in\left\{1,2\right\}\)

b) Đặt \(\left(7a+9b,3a+8b\right)=d\)

\(\Rightarrow\left\{{}\begin{matrix}7a+9b⋮d\\3a+8b⋮d\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}21a+27b⋮d\\21a+56b⋮d\end{matrix}\right.\) \(\Rightarrow29b⋮d\)

Lại có \(\left\{{}\begin{matrix}56a+72b⋮d\\27a+72b⋮d\end{matrix}\right.\Rightarrow29a⋮d\)

Mà \(\left(a,b\right)=1\) \(\Rightarrow d\in\left\{1,29\right\}\)

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