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 $|A|=3$, find the value of the real number $a$; (2) If $A=\varnothing$, find the range of values for the real number $a$.
First, we should interpret the set language in the question. $M \cap N$ means the solution set of the system of equations represented by the two sets $M$ and $N$. (1) is to find the value of the real number $a$ such that the system of equations has 3 solutions; (2) is to find the range of the real number $a$ such that the system of equations has no solutions.
Solving $\left\{\begin{array}{l}y=x^{2} \\ x^{2}+(y-a)^{2}=1\end{array} \Rightarrow y^{2}-(2 a-1) y+a^{2}-1=0\right.$
$\Delta=(2 a-1)^{2}-4\left(a^{2}-1\right)=5-4 a$.
When $a>\frac{5}{4}$, $\Delta<0$, the equation $(*)$ has two distinct real roots $y_{1}$ and $y_{2}$.
By $x^{2}=y \geqslant 0$, we get that the equation $(*)$ has one positive root and one root of 0 when the original system of equations has 3 solutions; the equation $(*)$ has two negative roots when the original system of equations has no solutions.
By $a^{2}-1=0 \Rightarrow a= \pm 1$. After verification, when $a=1$, the original system of equations has 3 solutions;
By $\left\{\begin{array}{l}\Delta>0 \\ 2 a-1<0\end{array} \Rightarrow a<\frac{1}{2}\right.$ and $a>\frac{5}{4}$.