34. (-$$\frac{5}{7}$$a³b²) ⋅ (-$$\frac{49}{25}$$a⁵b).

34. (-$$\frac{5}{7}$$a³b²) ⋅ (-$$\frac{49}{25}$$a⁵b) = (-$$\frac{5}{7}$$) ⋅ (-$$\frac{49}{25}$$) ⋅ a³ ⋅ a⁵ ⋅ b² ⋅ b = $$\frac{7}{5}$$a⁸b³ = 1$$\frac{2}{5}$$a⁸b³

Ответ: 1$$\frac{2}{5}$$a⁸b³