b, Ta có: 3.n+7\(⋮\) n

\(\Rightarrow\) 7\(⋮\) n (Vì 3.n\(⋮\) n)

\(\Rightarrow\) n\(\in\) Ư(7)=\(\left\{1;7\right\}\)

Ta có: (n^2+n+4)\(⋮\) n+1

\(\Rightarrow\)n(n+1)+4\(⋮\) n+1

\(\Rightarrow\) 4\(⋮\) n+1 (Vì n(n+1) đã chia hết cho 4)

\(\Rightarrow\) n+1\(\in\) U(4)\(\in\left\{1;2;4\right\}\)

\(\Rightarrow n\in\left\{0;1;3\right\}\)