$S_n=\left(\frac{n}{2}\right)[2a+(n-1\left)d\right]$
$\Rightarrow 0=\left(\frac{n}{2}\right)[64+(n-1\left)\right(-8\left)\right]$
$\Rightarrow 0=\left(\frac{n}{2}\right)[64-(n-1\left)8\right]$
$\Rightarrow$ n (36 – 4n) = 0
n = 0; n=9

As, n can not be equal to 0, so, the number of terms required is 9.

$S_n=\left[\frac{a(r^n-1)}{r-1}\right]$

In the question a = 3, r = 4, n = 5

$S_5=3\times \frac{(4^5-1)}{(4-1)}=1023$

$6,\_,\_,\_,96$

If the common ratio is r, $96=6(r{)}^4$

$r^4=16,r=2$

So, the geometric means are: 12, 24 and 48. 
a+b+c = 84

The series is a GP with common ratio$r=\left(\frac{-\frac{1}{6}}{\frac{1}{3}}\right)=-\frac{1}{2}$

Eight term$=a{r}^7=\frac{1}{3}\times {(-\frac{1}{2})}^7=-\frac{1}{384}$

$\frac{a}{r^2}$, $\frac{a}{r}$, $a$, $ar$, $a{r}^2$ are the first five terms.
Third term = 3
Product of first five terms = $a^5$ = 343.