(1) $a^3_n+\dfrac{a_n}{n}=1 $$\Rightarrow 0<a_n<1.$ ${{0=a^3_{n+1}}-a^3_n+\dfrac{a_{n+1}}{n+1}-\dfrac{a_n}{n}<a^3_{n+1}}-a^3_n+\dfrac{a_{n+1}}{n}-\dfrac{a_n}{n}$ $=(a_{n+1}-a_n)(a^2_{n+1}+a_{n+1}a_n+a^2_n+\dfrac{1}{n})$ $\Rightarrow a_{n+1}>a_n.$ (2) $a_n=\dfrac{1}{a^2_n+\dfrac{1}{n}}>\dfrac{1}{1+\dfrac{1}{n}}=\dfrac{n}{n+1},$ $\sum_{i=1}^{n}\frac{1}{(i+1)^2a_i}<\sum_{i=1}^{n}\frac{1}{i(i+1)}=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}<a_n.$