If the Second and fourth terms of a G.P. are 8 and 32 respectively, find the sum of the first nine terms. 2. A G.P. has its 3rd and 7th terms as 45 and 2025 respectively. What is the product of its 2nd and 8th terms?

The $n$-th term of a G.P. is given by $T_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio.
Given:
Second term ($T_2$) = 8
Fourth term ($T_4$) = 32

From the formula:

$$T_2 = ar^{2-1} = ar = 8 \quad \text{(Equation 1)}$$ $$T_4 = ar^{4-1} = ar^3 = 32 \quad \text{(Equation 2)}$$

Divide Equation 2 by Equation 1:

$$\frac{ar^3}{ar} = \frac{32}{8}$$ $$r^2 = 4$$ $$r = \sqrt{4}$$

Since the terms are positive, we take the positive root:

$$r = 2$$

Substitute $r=2$ into Equation 1:

$$a(2) = 8$$ $$a = \frac{8}{2}$$ $$a = 4$$

The sum of the first $n$ terms of a G.P. is given by $S_n = \frac{a(r^n - 1)}{r - 1}$ for $r \neq 1$.
Here, $a=4$, $r=2$, and $n=9$.

$$S_9 = \frac{4(2^9 - 1)}{2 - 1}$$ $$S_9 = \frac{4(512 - 1)}{1}$$ $$S_9 = 4(511)$$ $$S_9 = 2044$$

The sum of the first nine terms is $\boxed{2044}$.

2.

Given:
Third term ($T_3$) = 48
Seventh term ($T_7$) = $\frac{3}{4}$

From the formula $T_n = ar^{n-1}$:

$$T_3 = ar^{3-1} = ar^2 = 48 \quad \text{(Equation 1)}$$ $$T_7 = ar^{7-1} = ar^6 = \frac{3}{4} \quad \text{(Equation 2)}$$

Divide Equation 2 by Equation 1:

$$\frac{ar^6}{ar^2} = \frac{3/4}{48}$$ $$r^4 = \frac{3}{4 \times 48}$$ $$r^4 = \frac{3}{192}$$ $$r^4 = \frac{1}{64}$$

The 2nd term is $T_2 = ar$.
The 4th term is $T_4 = ar^3$.
The product of the 2nd and 4th terms is $T_2 \times T_4 = (ar)(ar^3) = a^2 r^4$.

From Equation 1, we have $ar^2 = 48$. We can square this to get $a^2 r^4$:

$$(ar^2)^2 = 48^2$$ $$a^2 r^4 = 2304$$

Alternatively, substitute $r^4 = \frac{1}{64}$ into $a^2 r^4$. First, find $a$:
From $ar^2 = 48$ and $r^4 = \frac{1}{64}$, we know $r^2 = \sqrt{\frac{1}{64}} = \frac{1}{8}$ (assuming $r^2$ is positive).

$$a \left(\frac{1}{8}\right) = 48$$ $$a = 48 \times 8 = 384$$

Now, calculate $a^2 r^4$:

$$a^2 r^4 = (384)^2 \times \frac{1}{64}$$ $$a^2 r^4 = 147456 \times \frac{1}{64}$$ $$a^2 r^4 = 2304$$

The product of its 2nd and 4th terms is $\boxed{2304}$.