1. If $16^{9 m}=a, 4^{37 n}=\frac{1}{a}$, then $(36 m+74 n-1)^{2000}$
Ni.1.1.
Since $a=16^{9 m}=\left(2^{4}\right)^{9 m}=2^{36 m}$, and $\frac{1}{a}=4^{37 n}=2^{74 n}$, then $1=a \times \frac{1}{a}=2^{36 m} \cdot 2^{74 n}=2^{36 m+74 n}$,
we have $36 m+74 n=0$.
Therefore, $(36 m+74 n-1)^{2000}=(-1)^{20000}=1$.