$$
\text { II. (40 points) Let } a_{i} \in(0,1](1 \leqslant i \leqslant 45) \text {. }
$$

Find the maximum value of $\lambda$ such that the inequality
$$
\sqrt{\frac{45}{\sum_{i=1}^{45} a_{i}}} \geqslant 1+\lambda \prod_{i=1}^{45}\left(1-a_{i}\right)
$$

holds.
Obviously, when $a_{i}$ are all 1, the equality in the problem's inequality holds.
It is easy to prove that when $a_{i}=\frac{1}{81}$, $\lambda=8\left(\frac{81}{80}\right)^{45}$.
Below is the proof:
$$
\sqrt{\frac{45}{\sum_{i=1}^{45} a_{i}}} \geqslant 1+8\left(\frac{81}{80}\right)^{45} \prod_{i=1}^{45}\left(1-a_{i}\right).
$$
Let $x>0$, and denote $45 x^{2}=\sum_{i=1}^{45} a_{i}$. Then $0<x \leqslant 1$.
By the AM-GM inequality, we have
$$
\begin{array}{l}
\prod_{i=1}^{45}\left(1-a_{i}\right) \leqslant\left(\frac{\sum_{i=1}^{45}\left(1-a_{i}\right)}{45}\right)^{45} \\
=\left(1-x^{2}\right)^{45}=(1-x)^{45}(1+x)^{45},
\end{array}
$$
The equality holds when $a_{i}$ are all equal.
Thus, it suffices to prove
$$
\begin{array}{l}
\frac{1}{x} \geqslant 1+8\left(\frac{81}{80}\right)^{45}(1-x)^{45}(1+x)^{45} \\
\Leftrightarrow\left(\frac{1-x}{x}\right)\left(1-8\left(\frac{81}{80}\right)^{45} x(1-x)^{44}(1+x)^{45}\right) \\
\quad \geqslant 0 .
\end{array}
$$
Since $1-x \geqslant 0$, it suffices to prove
$$
8\left(\frac{81}{80}\right)^{45} x(1-x)^{44}(1+x)^{45} \leqslant 1 \text {. }
$$
By the AM-GM inequality, we have
$$
\begin{array}{l}
8\left(\frac{81}{80}\right)^{45} x(1-x)^{44}(1+x)^{45} \\
=\left(\frac{9}{8}\right)^{90}(8 x)(1-x)^{44}\left(\frac{8}{10}(1+x)\right)^{45} \\
\leqslant\left(\frac{9}{8}\right)^{90}\left(\frac{8 x+44(1-x)+36(1+x)}{90}\right)^{90} \\
=\left(\frac{9}{8}\right)^{90}\left(\frac{80}{90}\right)^{90}=1 .
\end{array}
$$
In conclusion, $\lambda_{\text {max }}=8\left(\frac{81}{80}\right)^{45}$.