3. 已知$3x+5y-1= 0$,求$8^{x}\cdot 32^{y}$的值.
答案：解:$8^{x}\cdot 32^{y}=2^{3x}\cdot 2^{5y}=2^{3x+5y},$$\because 3x+5y-1=0,\therefore 3x+5y=1,\therefore$原式$=2^{1}=2.$
4. 计算：
(1)$5^{9}×0.2^{8}$； (2)$(\frac {3}{7})^{2024}×(-2\frac {1}{3})^{2025}$.
(1)$5^{9}×0.2^{8}$； (2)$(\frac {3}{7})^{2024}×(-2\frac {1}{3})^{2025}$.
答案：4.(1)$5^{9}×0.2^{8}=(5×0.2)^{8}×5=1×5=5.$(2)原式$=(\frac {3}{7})^{2024}×(-\frac {7}{3})^{2025}$$=-(\frac {3}{7})^{2024}×(\frac {7}{3})^{2025}$$=-(\frac {3}{7})^{2024}×(\frac {7}{3})^{2024}×\frac {7}{3},$$=-(\frac {3}{7}×\frac {7}{3})^{2024}×\frac {7}{3},$$=-\frac {7}{3}.$
解析：
(1)解：$5^{9}×0.2^{8}$
$=5×5^{8}×0.2^{8}$
$=5×(5×0.2)^{8}$
$=5×1^{8}$
$=5×1$
$=5$
(2)解：$(\frac {3}{7})^{2024}×(-2\frac {1}{3})^{2025}$
$=(\frac {3}{7})^{2024}×(-\frac {7}{3})^{2025}$
$=(\frac {3}{7})^{2024}×(-\frac {7}{3})^{2024}×(-\frac {7}{3})$
$=[\frac {3}{7}×(-\frac {7}{3})]^{2024}×(-\frac {7}{3})$
$=(-1)^{2024}×(-\frac {7}{3})$
$=1×(-\frac {7}{3})$
$=-\frac {7}{3}$
$=5×5^{8}×0.2^{8}$
$=5×(5×0.2)^{8}$
$=5×1^{8}$
$=5×1$
$=5$
(2)解：$(\frac {3}{7})^{2024}×(-2\frac {1}{3})^{2025}$
$=(\frac {3}{7})^{2024}×(-\frac {7}{3})^{2025}$
$=(\frac {3}{7})^{2024}×(-\frac {7}{3})^{2024}×(-\frac {7}{3})$
$=[\frac {3}{7}×(-\frac {7}{3})]^{2024}×(-\frac {7}{3})$
$=(-1)^{2024}×(-\frac {7}{3})$
$=1×(-\frac {7}{3})$
$=-\frac {7}{3}$
5. 若$a^{2n}= 3$,$b^{n}= \frac {1}{4}$,求$(-ab)^{2n}$的值.
答案：解:$\because a^{2n}=3,b^{n}=\frac {1}{4},\therefore (-ab)^{2n}=(ab)^{2n}=a^{2n}\cdot b^{2n}=$$a^{2n}\cdot (b^{n})^{2}=3×(\frac {1}{4})^{2}=3×\frac {1}{16}=\frac {3}{16}.$