37. Given that the integer part of the real number $\frac{2+\sqrt{2}}{2-\sqrt{2}}$ is $a$, and the fractional part is $1-b$. Then the value of $\frac{(b-1) \cdot(5-b)}{\sqrt{a^{2}-3^{2}}}$ is
Answer: -1

Solution:
$$
\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2 \sqrt{2}
$$

Since
$$
2 \sqrt{2}=\sqrt{8} \text {, }
$$

Therefore
$$
\begin{array}{c}
2<2 \sqrt{2}<3, \\
5<3+2 \sqrt{2}<6,
\end{array}
$$

We get
$$
a=5,1-b=(3+2 \sqrt{2})-5=2 \sqrt{2}-2, \quad b=3-2 \sqrt{2} \text {, }
$$

Thus
$$
\frac{(b-1) \cdot(5-b)}{\sqrt{a^{2}-3^{2}}}=\frac{(2-2 \sqrt{2})(2+2 \sqrt{2})}{\sqrt{5^{2}-3^{2}}}=-1 \text { . }
$$