II. (16 points) Given that $a$ and $b$ are real numbers, and $\mathrm{i}$ is the imaginary unit, the quadratic equation in $z$ is
$$
4 z^{2}+(2 a+\mathrm{i}) z-8 b(9 a+4)-2(a+2 b) \mathrm{i}=0
$$

has at least one real root. Find the maximum value of this real root.
Let $x$ be the real root of a known quadratic equation, then
$$
\begin{array}{l}
4 x^{2}+2 a x-8 b(9 a+4)+[x-2(a+2 b)] \mathrm{i}=0 \\
\Leftrightarrow\left\{\begin{array}{l}
4 x^{2}+2 a x-8 b(9 a+4)=0, \\
x-2(a+2 b)=0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
5 a^{2}+16\left(b-\frac{1}{4}\right)^{2}=1, \\
x=2(a+2 b) .
\end{array}\right.
\end{array}
$$

From equation (1), we can set $a=\frac{1}{\sqrt{5}} \cos \theta, b=\frac{1}{4} \sin \theta+\frac{1}{4}$, and substitute into equation (9) to get
$$
\begin{array}{l}
x=2\left(\frac{1}{\sqrt{5}} \cos \theta+\frac{1}{2} \sin \theta+\frac{1}{2}\right) \\
=\frac{3 \sqrt{5}}{5} \sin (\theta+\alpha)+1,
\end{array}
$$

where $\alpha$ is an acute angle, and $\sin \alpha=\frac{2}{3}$.
Since $\theta$ can take any real number, hence $x_{\max }=\frac{3 \sqrt{5}}{5}+1$.