(11) (12 points) Let $f(x)=x^{2}+m x+n$, if the inequality $|f(x)|>2$ has no solution in the interval $[1,5]$.
(1) Find the value of $f(1)-2 f(3)+f(5)$;
(2) Find all real pairs $(m, n)$.

11 Assume there exists such a pair of real numbers $(m, n)$, then $|f(x)| \leqslant 2$, for all $x \in [1,5]$. By taking $x=1,3,5$, we get
$$
\begin{array}{l}
|f(1)|=|1+m+n| \leqslant 2, \\
|f(3)|=|9+3 m+n| \leqslant 2, \\
|f(5)|=|25+5 m+n| \leqslant 2 .
\end{array}
$$

It is also easy to see that $f(1)-2 f(3)+f(5)=8$. Therefore,
$$
\begin{aligned}
8 & =|f(1)-2 f(3)+f(5)| \\
& \leqslant|f(1)|+2|f(3)|+|f(5)| \\
& \leqslant 2+2 \times 2+2=8 .
\end{aligned}
$$

Thus,
$$
\left\{\begin{array}{l}
f(1)=f(5)=2, \\
f(3)=-2,
\end{array}\right.
$$

which means
$$
\left\{\begin{array}{l}
1+m+n=25+5 m+n=2, \\
9+3 m+n=-2,
\end{array}\right.
$$

Solving these equations, we get
$$
m=-6, n=7 .
$$

When $m=-6$ and $n=7$, $f(x)=x^{2}-6 x+7$ satisfies $|f(x)| \leqslant 2$ for $x \in [1,5]$.
In conclusion, the required pair of real numbers $(m, n)=(-6,7)$.