One, (20 points) Given that the parabola $y=4 x^{2}+4 x+5$ and the line $y=8 m x+8 m$ have only one common point. Find the value of $m^{36} +\frac{1155}{m^{12}}$.
1. Eliminating $y$ yields
$$
4 x^{2}+(4-8 m) x+5-8 m=0 \text {. }
$$

By the problem's condition, the equation has two equal real roots, hence
$$
\begin{array}{l}
(4-8 m)^{2}-4 \times 4(5-8 m)=0 \\
\Rightarrow m^{2}+m-1=0 \Rightarrow m-\frac{1}{m}=-1 . \\
\text { Also, } m^{3}-\frac{1}{m^{3}}=\left(m-\frac{1}{m}\right)^{3}+3\left(m-\frac{1}{m}\right)=-2 \\
\Rightarrow m^{6}+\frac{1}{m^{6}}=6 \\
\Rightarrow m^{12}+\frac{1}{m^{12}}=34, m^{24}+\frac{1}{m^{24}}=1154 \\
\Rightarrow m^{36}+\frac{1}{m^{12}}=1154 m^{12}, \\
1154 m^{12}+\frac{1154}{m^{12}}=34 \times 1154 .
\end{array}
$$

Adding the above two equations yields
$$
m^{36}+\frac{1155}{m^{12}}=34 \times 1154=39236 .
$$