2) a) (3x⁶y⁴)⋅\(-\frac{1}{81}xy²\); б) \(-\frac{2}{3}ab⁵\)⋅18a⁵b.

2) a) $$(3x^6y^4) \cdot (-\frac{1}{81}xy^2) = -\frac{3}{81}x^7y^6 = -\frac{1}{27}x^7y^6$$.

б) $$(-\frac{2}{3}ab^5) \cdot 18a^5b = -\frac{2}{3} \cdot 18 a^6b^6 = -12a^6b^6$$.

Ответ: $$\frac{-1}{27}x^7y^6; -12a^6b^6$$