Use the iterative formula xn+1 = sin⁰¹(2/(xn+2)) with initial value x¹ = 0.75, to determine this root correct to 2 decimal places. Give the result of each iteration to decimal places.

The iterations converge to 0.7970....

The root correct to 2 decimal places is $\boxed{0.80}$.

Question 14.i) Simplify and deduce:
Simplify $(\sqrt{1+x}+\sqrt{1-x})(\sqrt{1+x}-\sqrt{1-x})$.

Let $a = \sqrt{1+x}$ and $b = \sqrt{1-x}$.

The simplified expression is $\boxed{2x}$.

From the simplification, we have:

Divide both sides by $(\sqrt{1+x}+\sqrt{1-x})$:

Now, divide both sides by $2x$:

This deduction is shown.

Question 14.ii) Obtain the expansion of $\frac{1}{\sqrt{1+x}+\sqrt{1-x}}$ in ascending powers of $x$, up to and including the term in $x^2$:

$\frac{1}{\sqrt{1+x}+\sqrt{1-x}} = \frac{\sqrt{1+x}-\sqrt{1-x}}{2x}$.

For $\sqrt{1+x} = (1+x)^{1/2}$:

For $\sqrt{1-x} = (1-x)^{1/2}$:

Now divide by $2x$:

The expansion is $\boxed{\frac{1}{2} + \frac{1}{16}x^2}$.

Question 15a) i) Express $z$ in the form $x+iy$:
Given $z = \frac{4-3i}{1-2i}$.

The conjugate of $1-2i$ is $1+2i$.

Denominator: $(1-2i)(1+2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 1+4 = 5$.
Numerator: $(4-3i)(1+2i) = 4(1) + 4(2i) - 3i(1) - 3i(2i) = 4 + 8i - 3i - 6i^2 = 4 + 5i - 6(-1) = 4 + 5i + 6 = 10 + 5i$.

So, $z = \boxed{2+i}$.

Question 15a) ii) Find the modulus and argument of $z$:
Given $z = 2+i$.

The modulus $|z|$ is given by $\sqrt{x^2+y^2}$.

The argument $\arg(z)$ is given by $\tan^{-1}\left(\frac{y}{x}\right)$. Since $x=2$ and $y=1$ are both positive, $z$ is in the first quadrant.

Using a calculator, $\tan^{-1}(0.5) \approx 0.4636$ radians (or $26.57^\circ$).
The modulus is $\boxed{\sqrt{5}}$ and the argument is $\boxed{\tan^{-1}(0.5) \text{ radians}}$.

Question 15b) Find the two square roots of the complex number $5-12i$:
Let the square roots be $x+iy$, where $x$ and $y$ are real.

Equating real parts:

Equating imaginary parts:

From (2), $y = -\frac{6}{x}$. Substitute this into (1):

Multiply by $x^2$ (assuming $x \neq 0$):

This is a quadratic equation in $x^2$. Let $u = x^2$.

Factorize the quadratic:

So, $u=9$ or $u=-4$.
Since $u=x^2$ and $x$ is real, $x^2$ must be non-negative. Thus, $x^2 = 9$.

If $x=3$, from $xy=-6$, $3y=-6 \implies y=-2$.
So, one square root is $3-2i$.
If $x=-3$, from $xy=-6$, $-3y=-6 \implies y=2$.
So, the other square root is $-3+2i$.
The two square roots are $\boxed{3-2i \text{ and } -3+2i}$.

Question 16) Prove by induction that $5^{2n} - 3^{2n}$ is a multiple of 8, for $n \ge 1$:

For $n=1$, we need to show that $5^{2(1)} - 3^{2(1)}$ is a multiple of 8.

Since $16 = 8 \times 2$, it is a multiple of 8.
The base case holds.

Assume that for some positive integer $k \ge 1$, $5^{2k} - 3^{2k}$ is a multiple of 8.
This means $5^{2k} - 3^{2k} = 8m$ for some integer $m$.
From this, we can write $5^{2k} = 3^{2k} + 8m$.

We need to show that $5^{2(k+1)} - 3^{2(k+1)}$ is a multiple of 8.

Substitute $5^{2k} = 3^{2k} + 8m$ from the inductive hypothesis:

Group terms with $3^{2k}$:

Factor out 8:

Since $k \ge 1$, $3^{2k}$ is an integer, and $m$ is an integer, the expression $(2 \cdot 3^{2k} + 25m)$ is an integer.
Therefore, $5^{2(k+1)} - 3^{2(k+1)}$ is a multiple of 8.

By the principle of mathematical induction, $5^{2n} - 3^{2n}$ is a multiple of 8 for all integers $n \ge 1$.