If $\frac{(a + b)^3}{(a - b)^3} = \frac{64}{27}$

(a) Solving,

$$\begin{aligned}\Rightarrow \frac{(a + b)^3}{(a - b)^3} = \frac{64}{27} \\[1em] \Rightarrow \frac{(a + b)^3}{(a - b)^3} = \frac{4^3}{3^3} \\[1em] \Rightarrow \Big(\frac{a + b}{a - b}\Big)^3 = \Big(\frac{4}{3}\Big)^3 \\[1em] \Rightarrow \frac{a + b}{a - b} = \frac{4}{3} \\[1em]\end{aligned}$$

Hence, $\frac{a + b}{a - b} = \frac{4}{3}.$

(b) Solving further,

$$\begin{aligned}\Rightarrow 3(a + b) = 4(a - b) \\[1em] \Rightarrow 3a + 3b = 4a - 4b \\[1em] \Rightarrow 4a - 3a = 3b + 4b \\[1em] \Rightarrow a = 7b \\[1em] \Rightarrow \frac{a}{b} = \frac{7}{1} \\[1em] \Rightarrow a : b = 7 : 1.\end{aligned}$$

Hence, a : b = 7 : 1.