Question 7: Given $M=\left\{(x, y) \mid y=x^{2}\right\}, N=\left\{(x, y) \mid x^{2}+(y-a)^{2}=1\right\}, A=M \cap N$.
(1) If $|\mathrm{A}|=3$, find the range of real number $a$;
(2) If $A=\emptyset$, find the range of real number $a$.
Question 7, Solution: Solve the system of equations:
$$
\left\{\begin{array}{c}
y = x^2 \\
x^2 + (y - a)^2 = 1
\end{array} \Leftrightarrow \left\{\begin{array}{c}
y^2 - (2a - 1)y + a^2 - 1 = 0 \\
x = \pm \sqrt{y}
\end{array}\right.\right.
$$

Consider the quadratic equation:
$$
y^2 - (2a - 1)y + a^2 - 1 = 0
$$

Its discriminant is $\Delta = (2a - 1)^2 - 4(a^2 - 1) = 5 - 4a$.
(1) The condition $|\mathrm{A}| = 3$ is equivalent to the equation $(*)$ having exactly one positive root and the other root being 0, thus $a \in \{\pm 1\}$. Upon verification, $a = 1$ satisfies the condition.
(2) The condition $A = \emptyset$ is equivalent to $\Delta < 0$:
$$
\begin{array}{c}
\Delta < 0 \\
5 - 4a < 0 \\
a > \frac{5}{4}
\end{array} \Leftrightarrow \left\{\begin{array}{c}
a \in \left(-\infty, \frac{5}{4}\right] \\
a \in \left(-\infty, \frac{1}{2}\right] \\
a \in (-\infty, -1) \cup (1, +\infty)
\end{array} \Leftrightarrow a \in (-\infty, -1)\right.\right.
$$

Combining both cases, we find that the range of $a$ is $(-\infty, -1) \cup \left(\frac{5}{4}, +\infty\right)$.