Five. (20 points) Let
$$
f(x)=\ln x-\frac{1}{2} a x^{2}-2 x \quad (a<0),
$$

and the function $f(x)$ has a monotonically decreasing interval.
(1) Find the range of values for $a$;
(2) If for any value of $a$ that satisfies the condition, $f(x)<b$ holds for all $x \in (0,1]$, find the range of values for the real number $b$.
(1) From the given, we have
$$
f^{\prime}(x)=\frac{1}{x}-a x-2=-\frac{a x^{2}+2 x-1}{x}(x>0) \text {. }
$$

Since the function $f(x)$ has a monotonically decreasing interval, the inequality $f^{\prime}(x) \leqslant 0$, i.e., $a x^{2}+2 x-1 \geqslant 0$ has solutions on $(0,+\infty)$.

Noting that, $a0$.
Thus, $\Delta=4+4 a \geqslant 0 \Rightarrow a \geqslant-1$. Therefore, the range of $a$ is $[-1,0)$.
(2) According to the problem, for any $a \in[-1,0)$, we have $b>\left(\ln x-\frac{1}{2} a x^{2}-2 x\right)_{\max }$.
Noting that the function
$$
h(a)=\ln x-\frac{1}{2} a x^{2}-2 x
$$

is monotonically decreasing on $a \in[-1,0)$. Thus,
$$
h(a)_{\max }=h(-1)=\frac{1}{2} x^{2}-2 x+\ln x .
$$

Therefore, $b>\left(\frac{1}{2} x^{2}-2 x+\ln x\right)_{\max }$.
Let $\varphi(x)=\frac{1}{2} x^{2}-2 x+\ln x$. Then
$$
\varphi^{\prime}(x)=x-2+\frac{1}{x}=\frac{(x-1)^{2}}{x} \geqslant 0 \text {. }
$$

Thus, the function $\varphi(x)$ is monotonically increasing on $(0,1]$.
Hence, $\varphi(x)_{\max }=\varphi(1)=-\frac{3}{2}$.
Therefore, $b>-\frac{3}{2}$, i.e., $b \in\left(-\frac{3}{2},+\infty\right)$.