$1.6 \frac{\left(\frac{1}{a}+\frac{1}{b}-\frac{2 c}{a b}\right)(a+b+2 c)}{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{a b}-\frac{4 c^{2}}{a^{2} b^{2}}} ; a=7,4, b=\frac{5}{37}$.
1.6 Perform the following transformations in the numerator and denominator:

$$
\begin{aligned}
& \frac{1}{a}+\frac{1}{b}-\frac{2 c}{a b}=\frac{b+a-2 c}{a b} \\
& \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{a b}-\frac{4 c^{2}}{a^{2} b^{2}}=\frac{b^{2}+a^{2}+2 a b-4 c^{2}}{a^{2} b^{2}}= \\
& =\frac{(b+a)^{2}-(2 c)^{2}}{a^{2} b^{2}}=\frac{(b+a-2 c)(b+a+2 c)}{a^{2} b^{2}}
\end{aligned}
$$

Thus,

$$
\frac{(b+a-2 c)(b+a+2 c)}{a b} : \frac{(b+a-2 c)(b+a+2 c)}{a^{2} b^{2}}=a b
$$

When $a=7.4=\frac{37}{5}, b=\frac{5}{37}$, we get $a b=1$.

Answer: 1.