(1) From $a_{n}^{3}+\frac{a_{n}}{n}=1$, we get $0 < a_{n} < 1$, so,
$$a_{n+1}-a_{n}>0,$$

which means $a_{n+1}>a_{n}$.
(2) Since $a_{n}\left(a_{n}^{2}+\frac{1}{n}\right)=1$, we have,
$$a_{n}=\frac{1}{a_{n}^{2}+\frac{1}{n}}>\frac{1}{1+\frac{1}{n}}=\frac{n}{n+1} \text {. }$$

Thus, $\frac{1}{(n+1)^{2} a_{n}}<\frac{1}{n(n+1)}$.
Therefore, $\sum_{i=1}^{n} \frac{1}{(i+1)^{2} a_{i}}<\sum_{i=1}^{n} \frac{1}{i(i+1)}$
$$=\sum_{i=1}^{n}\left(\frac{1}{i}-\frac{1}{i+1}\right)=1-\frac{1}{n+1}=\frac{n}{n+1}<a_{n} \text {. }$$