$\frac{{{n}^{2}}+5}{n+2}=\frac{n\left( n+2 \right)-2n+5}{n+2}=\frac{n\left( n+2 \right)-2\left( n+2 \right)+9}{n+2}=n-2+\frac{9}{n+2}$
Do đó để $\left( {{n}^{2}}+5 \right)\vdots \left( n+2 \right)$ thì $9\vdots \left( n+2 \right).$
Do $n\in \mathbb{N}$ nên $n+2\ge 2.$
Do đó $\left( n+2 \right)\in \left\{ 3,9 \right\}$, suy ra $n\in \left\{ 1,7 \right\}.$
Vậy $n\in \left\{ 1,7 \right\}$ thì $\left( {{n}^{2}}+5 \right)\vdots \left( n+2 \right).$

Tôi nghĩ n€{1;7}