12. A. Do there exist prime numbers $p$ and $q$ such that the quadratic equation $p x^{2}-q x+p=0$ has rational roots?
12.A. Suppose the equation has a rational root. Then the discriminant is a perfect square. Let
$$
\Delta=q^{2}-4 p^{2}=n^{2},
$$

where $n$ is a non-negative integer. Then
$$
(q-n)(q+n)=4 p^{2} \text {. }
$$

Since $1 \leqslant q-n \leqslant q+n$, and $q-n$ and $q+n$ have the same parity, they must both be even. Therefore, there are the following possible cases:
(1) $\left\{\begin{array}{l}q-n=2, \\ q+n=2 p^{2}\end{array} \Rightarrow q=p^{2}+1\right.$;
(2) $\left\{\begin{array}{l}q-n=4, \\ q+n=p^{2}\end{array} \Rightarrow q=2+\frac{p^{2}}{2}\right.$;
(3) $\left\{\begin{array}{l}q-n=p, \\ q+n=4 p\end{array} \Rightarrow q=\frac{5 p}{2}\right.$;
(4) $\left\{\begin{array}{l}q-n=2 p \\ q+n=2 p\end{array} \Rightarrow q=2 p\right.$;
(5) $\left\{\begin{array}{l}q-n=p^{2} \\ q+n=4\end{array}, \Rightarrow q=2+\frac{p^{2}}{2}\right.$.
For cases (1) and (3), $p=2$, thus, $q=5$;
For cases (2) and (5), $p=2$, thus, $q=4$ (which does not meet the condition, discard);

For case (4), $q$ is a composite number (which does not meet the condition, discard).

Furthermore, when $p=2, q=5$, the equation is $2 x^{2}-5 x+2=0$, its roots are $x_{1}=\frac{1}{2}, x_{2}=2$, which are both rational numbers.
In summary, there exists a prime number that satisfies the given conditions.