An infinite geometric progression has common ratio r with |r| < 1…

(a), (b) Squaring the first equation and dividing by the second: $\dfrac{a^2}{(1 - r)^2} \cdot \dfrac{(1 - r)(1 + r)}{a^2} = \dfrac{400}{100}$, so $\dfrac{1 + r}{1 - r} = 4$ and $r = \dfrac{3}{5}$. Then $a = 20(1 - r) = 8$.

(c) The cubes form a GP with first term $8^3 = 512$ and ratio $r^3 = \dfrac{27}{125}$, so their sum is $\dfrac{512}{1 - \frac{27}{125}} = \dfrac{32000}{49} \approx 653.06$.

(d) What is left after $n$ terms is $20r^n = 20\left(\tfrac35\right)^n$. We need $20\left(\tfrac35\right)^n < 0.01$: for $n = 15 - 1$ it is about $0.01567$ and for $n = 15$ about $0.00940$. So the least $n$ is $15$.

(e) The terms in even places are $ar, ar^3, ar^5, \dots$: a GP with first term $ar$ and ratio $r^2 = \dfrac{9}{25}$. Their sum is $\dfrac{ar}{1 - r^2} = \dfrac{8 \times \frac35}{1 - \frac{9}{25}} = \dfrac{15}{2} = 7.50$.