(1) When a person replaces the two middle digits $\overline{00}$ in 2009 with the two-digit numbers $\overline{a b}$ and $c \bar{d}$, they exactly get perfect squares: $\overline{2 a b 9}=n^{2}, \overline{2 c d 9}=m^{2},(m>n, m, n \in$ $\mathbf{N})$, then the array $(m+n, \overline{a b}+\bar{c} \bar{d})=$ $\qquad$ .
(1) $(100,100)$ Hint: Note that, for an integer $k$, if the last digit of $k^{2}$ is 9, then the last digit of $k$ must be 3 or 7. It is easy to see that $44^{2}\overline{2 c d 9}$, so $44<n<m<55$. Therefore, if $m$ and $n$ are to satisfy the condition, the only possibility is: $n=47, m=53$. Since $47^{2}=2209, 53^{2}=2809$, we have $\overline{a b}=20, \overline{c d}=80, n=47, m=53, (m+n, \overline{a b}+\overline{c d})=$ $(100,100)$.