б) $$(3a^2b^3c^2)^3 : (-2a^2cb^4)^2 = \frac{(3a^2b^3c^2)^3}{(-2a^2cb^4)^2} = \frac{3^3a^{2 \cdot 3}b^{3 \cdot 3}c^{2 \cdot 3}}{(-2)^2a^{2 \cdot 2}c^2b^{4 \cdot 2}} = \frac{27a^6b^9c^6}{4a^4c^2b^8} = \frac{27}{4} \cdot a^{6-4} \cdot b^{9-8} \cdot c^{6-2} = \frac{27}{4}a^2bc^4$$
Ответ: $$\frac{27}{4}a^2bc^4$$