Repeatedly apply the calculation rules of powers. This leads to: $$\begin{aligned}(-4a^2b^3)^{3} (-a^4b^5)^{2}&=(-4)^{3}\cdot(a^{2})^{3}\cdot (b^{3})^{3}\cdot (-1)^{2}\cdot(a^{4})^{2}\cdot (b^{5})^{2}\\ &\phantom{abcdefghirstuvwxyz}\blue{\text{application of }(x\cdot y)^r = x^r\cdot y^r}\\ &= -64\cdot a^{(2\times3)}\cdot b^{(3\times3)}\cdot 1\cdot a^{(4\times2)}\cdot b^{(5\times2)}\\ &\phantom{abcdefghirstuvwxyz}\blue{\text{application of }(x^r)^s = x^{r\cdot s}}\\ &=-64\,a^6b^9\cdot \,a^8b^{10}\\ &\phantom{abcdefghirstuvwxyz}\blue{\text{calculation of exponents}}\\ &=(-64\times1)\cdot{a}^{(6+8)}{b}^{(9+10)}\\&\phantom{abcdefghirstuvwxyz}\blue{\text{application of }(x\cdot y)^r = x^r\cdot y^r} \\ &=-64\,{a}^{14}{b}^{19}\\ &\phantom{abcdefghirstuvwxyz}\blue{\text{calculation of exponents}}\end{aligned}$$