Solution:

a) We have $a_{2}=a_{1}+\frac{1}{a_{1}} \geq 2$. If $a_{n} \geq n$, then
$$a_{n+1}-n-1=a_{n}+\frac{n}{a_{n}}-n-1=\frac{\left(a_{n}-1\right)\left(a_{n}-n\right)}{a_{n}} \geq 0$$
and the assertion follows by induction.

b) Let $n \geq 2$. It follows from a) that $a_{n+1} \leq a_{n}+1$. Then $a_{n} \leq a_{2}+n-2$, whence $1 \leq \frac{a_{n}}{n} \leq 1+\frac{a_{2}-2}{n}$. Therefore the sequence $\left(\frac{a_{n}}{n}\right)_{n \geq 1}$ is convergent and its limit equals $1$.

Remark. One can prove the stronger statement that $\lim _{n \rightarrow \infty}\left(a_{n}-n\right)=0$.