Without actually calculating the cubes,

(i) Given, $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$ or $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}+\left(-\frac{5}{6}\right)^{3}$

Here, we see that, $\frac{1}{2}+\frac{1}{3}-\frac{5}{6}=\frac{3+2-5}{6}=\frac{5-5}{6}=0$

$\therefore \quad\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}=3 \times \frac{1}{2} \times \frac{1}{3} \times\left(-\frac{5}{6}\right)=-\frac{5}{12}$

[using identity, if $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$ ]

(ii) Given, $(02)^{3}-(0.3)^{3}+(0.1)^{3}$ or $(02)^{3}+(-0.3)^{3}+(0.1)^{3}$

Here, we see that, $02-0.3+0.1=0.3-0.3=0$

$\therefore \quad(02)^{3}+(-0.3)^{3}+(0.1)^{3}=3 \times(02) \times(-0.3) \times(0.1)$

[using identity, if $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$ ]

$=-0.6 \times 0.03=-0.018$