3. Given that $m$ is an integer, the equation
$$
x^{2}-m x+3-n=0
$$

has two distinct real roots, the equation
$$
x^{2}+(6-m) x+7-n=0
$$

has two equal real roots, and the equation
$$
x^{2}+(4-m) x+5-n=0
$$

has no real roots.
$$
\text { Then }(m-n)^{2013}=
$$
3. According to the problem, we have
$$
\left\{\begin{array}{l}
m^{2}-4(3-n)>0, \\
(6-m)^{2}-4(7-n)=0, \\
(4-m)^{2}-4(5-n)\frac{5}{3}, m<3 \text {. }
\end{array}\right.
$$

Then $m=2$. Consequently, $n=3$.
Therefore, $(m-n)^{2013}=-1$.