The sum of the first and second terms of a geometric progression is -5, and the sum of the fourth and fifth terms is 40. Find the first term and the common ratio.

The $n$-th term of a geometric progression is $a_n = ar^{n-1}$.
The sum of the first and second terms is -5:

$$a_1 + a_2 = -5$$ $$a + ar = -5$$ $$a(1+r) = -5 \quad \text{(Equation 1)}$$

The sum of the fourth and fifth terms is 40:

$$a_4 + a_5 = 40$$ $$ar^3 + ar^4 = 40$$ $$ar^3(1+r) = 40 \quad \text{(Equation 2)}$$

Divide Equation 2 by Equation 1:

$$\frac{ar^3(1+r)}{a(1+r)} = \frac{40}{-5}$$

Assuming $a \neq 0$ and $1+r \neq 0$:

$$r^3 = -8$$ $$r = \sqrt[3]{-8}$$ $$r = -2$$

Substitute $r=-2$ into Equation 1:

$$a(1+(-2)) = -5$$ $$a(1-2) = -5$$ $$a(-1) = -5$$ $$a = 5$$

The first term is $a=5$ and the common ratio is $r=-2$.

$$\text{First term } = \boxed{5}$$ $$\text{Common ratio } = \boxed{-2}$$

8. (ii)

The general term $T_{k+1}$ in the expansion of $(A+B)^n$ is given by $T_{k+1} = \binom{n}{k} A^{n-k} B^k$.
For $\left(x + \frac{2}{x^2}\right)^{15}$, we have $A=x$, $B=\frac{2}{x^2}$, and $n=15$.

$$T_{k+1} = \binom{15}{k} (x)^{15-k} \left(\frac{2}{x^2}\right)^k$$

$$T_{k+1} = \binom{15}{k} x^{15-k} \frac{2^k}{(x^2)^k}$$ $$T_{k+1} = \binom{15}{k} x^{15-k} \frac{2^k}{x^{2k}}$$ $$T_{k+1} = \binom{15}{k} 2^k x^{15-k-2k}$$ $$T_{k+1} = \binom{15}{k} 2^k x^{15-3k}$$

For the term to be independent of $x$, the exponent of $x$ must be 0.

$$15 - 3k = 0$$ $$3k = 15$$ $$k = 5$$

The term independent of $x$ is $T_{5+1} = T_6$:

$$T_6 = \binom{15}{5} 2^5 x^{15-3(5)}$$ $$T_6 = \binom{15}{5} 2^5 x^0$$ $$T_6 = \binom{15}{5} 2^5$$

Calculate $\binom{15}{5}$:

$$\binom{15}{5} = \frac{15!}{5!(15-5)!} = \frac{15!}{5!10!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}$$ $$\binom{15}{5} = \frac{360360}{120} = 3003$$

Calculate $2^5$:

$$2^5 = 32$$

Multiply the values:

$$T_6 = 3003 \times 32$$ $$T_6 = 96096$$

The term independent of $x$ is $\boxed{96096}$.