(9) Given the quadratic function $y=f(x)=x^{2}+b x+c$ whose graph passes through the point $(1,13)$, and the function $y=f\left(x-\frac{1}{2}\right)$ is an even function.
(1) Find the analytical expression of $f(x)$;
(2) Does there exist a point on the graph of the function $y=f(x)$ such that its x-coordinate is a positive integer and its y-coordinate is a perfect square? If such points exist, find their coordinates; if not, explain why.
(9) (1) Since the function $y=f\left(x-\frac{1}{2}\right)$ is an even function, the axis of symmetry of the quadratic function $f(x)=x^{2}+b x+c$ is $x=-\frac{1}{2}$, hence $b=1$.

Since the graph of the quadratic function $f(x)=x^{2}+b x+c$ passes through the point $(1,13)$, we have $1+b+c=13$, hence $c=11$.
Therefore, the expression for $f(x)$ is $f(x)=x^{2}+x+11$.
(2) If there exists a point on the graph of the function $y=f(x)$ that meets the requirements, let it be $P$ $\left(m, n^{2}\right)$, where $m$ is a positive integer and $n$ is a natural number, then
$$
m^{2}+m+11=n^{2},
$$

thus
$$
4 n^{2}-(2 m+1)^{2}=43,
$$

which means
$$
[2 n+(2 m+1)][2 n-(2 m+1)]=43 .
$$

Noting that 43 is a prime number, and
$$
\begin{array}{c}
2 n+(2 m+1)>2 n-(2 m+1), \\
2 n+(2 m+1)>0,
\end{array}
$$

we have $\left\{\begin{array}{l}2 n+(2 m+1)=43, \\ 2 n-(2 m+1)=1,\end{array}\right.$ solving this gives $\left\{\begin{array}{l}m=10, \\ n=11 .\end{array}\right.$
Therefore, there exists a point on the graph of the function $y=f(x)$ that meets the requirements, and its coordinates are $(10,121)$.