Question 8: Let set $A=\left\{(x, y) \mid y^{2}-x-1=0\right\}$, set $B=\left\{(x, y) \mid 4 x^{2}+2 x-2 y+5=\right.$ $0\}$, and set $\mathrm{C}=\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{y}=\mathrm{kx}+\mathrm{b}\}$. Try to find all non-negative integers $\mathrm{k}$ and $\mathrm{b}$, such that $(\mathrm{A} \cup \mathrm{B}) \cap \mathrm{C}=\emptyset$.
Question 8, Answer: Since $(A \cup B) \cap C=\emptyset$, it follows that $A \cap C=B \cap C=\varnothing$.
Solving the system $\left\{\begin{array}{c}y^{2}-x-1=0 \\ y=k x+b\end{array}\right.$, we get $k^{2} x^{2}+(2 b k-1) x+\left(b^{2}-1\right)=0$, hence its discriminant
$$
\begin{array}{l}
\Delta_{1}=(2 \mathrm{bk}-1)^{2}-4 \mathrm{k}^{2}\left(\mathrm{~b}^{2}-1\right)=4 \mathrm{k}^{2}-4 \mathrm{~kb}+10 \\ 20-8 b>0\end{array}\right.$, so $b=2$. Consequently, $k=1$.
Upon verification, when $(k, b)=(1,2)$, (1) and (2) both hold true, hence the solution satisfying the conditions is $(k, b)=(1,2)$.