The sum of the second and third terms of a G.P. is six times the fourth term. If the second term is 8, calculate (a) the two possible values of the common ratio, and (b) the first 5 terms of the sequence.

Since $T_2 = ar = 8$, we know $ar \neq 0$. We can divide the entire equation by $ar$:

Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=6$, $b=-1$, $c=-1$:

The two possible values for $r$ are:

The two possible values of the common ratio are $\boxed{\frac{1}{2} \text{ and } -\frac{1}{3}}$.

$T_1 = a = 16$
$T_2 = ar = 16 \times \frac{1}{2} = 8$
$T_3 = ar^2 = 16 \times \left(\frac{1}{2}\right)^2 = 16 \times \frac{1}{4} = 4$
$T_4 = ar^3 = 16 \times \left(\frac{1}{2}\right)^3 = 16 \times \frac{1}{8} = 2$
$T_5 = ar^4 = 16 \times \left(\frac{1}{2}\right)^4 = 16 \times \frac{1}{16} = 1$
The first 5 terms are $\boxed{16, 8, 4, 2, 1}$.

Case 2: When $r = -\frac{1}{3}$

$T_1 = a = -24$
$T_2 = ar = -24 \times \left(-\frac{1}{3}\right) = 8$
$T_3 = ar^2 = -24 \times \left(-\frac{1}{3}\right)^2 = -24 \times \frac{1}{9} = -\frac{24}{9} = -\frac{8}{3}$
$T_4 = ar^3 = -24 \times \left(-\frac{1}{3}\right)^3 = -24 \times \left(-\frac{1}{27}\right) = \frac{24}{27} = \frac{8}{9}$
$T_5 = ar^4 = -24 \times \left(-\frac{1}{3}\right)^4 = -24 \times \frac{1}{81} = -\frac{24}{81} = -\frac{8}{27}$
The first 5 terms are $\boxed{-24, 8, -\frac{8}{3}, \frac{8}{9}, -\frac{8}{27}}$.