A geometric progression has positive terms and is increasing…

(a) $\dfrac{ar + ar^3}{ar^2} = \dfrac{30}{12}$ gives $\dfrac{1 + r^2}{r} = \dfrac52$, that is $2r^2 - 5r + 2 = 0$, so $r = 2$ or $r = \dfrac{1}{2}$. The progression is increasing, so $r = 2$.

(b) $a = \dfrac{12}{r^2} = 3$. The GP is $3, 6, 12, 24, \dots$; check: $6 + 24 = 30$.

(c) $S_8 = \dfrac{3(2^8 - 1)}{2 - 1} = 3 \times 255 = 765$.

(d) $3(2^n - 1) > 10\,000$ needs $2^n > \frac{10003}{3}$. Since $2^{11} = 2048$ and $2^{12} = 4096$, the least $n$ is $12$. Check: $S_{11} = 6141$ and $S_{12} = 12285$.

(e) The reciprocals $\dfrac13, \dfrac16, \dfrac{1}{12}, \dots$ form a GP with ratio $\dfrac12$: the sum is $\dfrac{\frac13\left(1 - \frac{1}{256}\right)}{1 - \frac12} = \frac{85}{128} \approx 0.66$.