Here are the solutions for questions 2, 3, and 4. 2a) Solve the following system of equations: x + 2y = 8, -x + y = -2. 2b) Find the value(s) of x in (2^x)2 - 5(2^x) + 4 = 0. 3) A firm buys a machine

The solution is $x=4$ and $y=2$.
The final answer is $\boxed{x=4, y=2}$.

2b) Find the value(s) of $x$ in $(2^x)^2 - 5(2^x) + 4 = 0$.

Substitute $u$ into the equation to form a quadratic equation.

This gives two possible values for $u$: $u=1$ or $u=4$.

For $u=1$:

For $u=4$:

The values of $x$ are 0 and 2.
The final answer is $\boxed{x=0, 2}$.

3) A firm buys a machine for N32,500. It depreciates every year at the rate of 10%. Find its value after 4 years.

$P_0 = N32,500$
$r = 10\% = 0.10$
$n = 4$ years

The value of the machine after 4 years is N21,323.25.
The final answer is $\boxed{\text{N}21,323.25}$.

4a) Find the coefficient of $x^8$ in the expansion of $(x^2 + \frac{2}{x})^{11}$.

Here, $a = x^2$, $b = \frac{2}{x}$, and $n = 11$.

Since $r$ must be a non-negative integer for a term to exist in the binomial expansion, and $\frac{14}{3}$ is not an integer, there is no term containing $x^8$ in this expansion.
The final answer is $\boxed{0}$.

4b) Simplify $\log_b x^2 - \log_b (x-3) = \log_b (x+4)$.

Multiply both sides by $(x-3)$.

Subtract $x^2$ from both sides:

For $\log_b A$, $A$ must be positive.
$x^2 > 0 \implies x \neq 0$
$x-3 > 0 \implies x > 3$
$x+4 > 0 \implies x > -4$
The solution $x=12$ satisfies all these conditions.
The final answer is $\boxed{x=12}$.