The function h is defined as follows. $h : x \mapsto \ln(x - k + 1),\ x \in \mathbb{R},\ x \ge k,$ where $k < 0$.

1. Differentiate $f(x)$ using the chain rule.

$f(x) = e^{\tan^{-1}\left(\frac{x}{2}\right)}$
$f'(x) = e^{\tan^{-1}\left(\frac{x}{2}\right)} \cdot \frac{d}{dx}\left[\tan^{-1}\left(\frac{x}{2}\right)\right]$

2. Evaluate the derivative of the inverse tangent function.

$\frac{d}{dx}\left[\tan^{-1}\left(\frac{x}{2}\right)\right] = \frac{1}{1 + \left(\frac{x}{2}\right)^2} \cdot \frac{1}{2}$
$= \frac{1}{1 + \frac{x^2}{4}} \cdot \frac{1}{2}$
$= \frac{4}{4 + x^2} \cdot \frac{1}{2}$
$= \frac{2}{4 + x^2}$

3. Substitute back and rearrange to obtain the required equation.

$f'(x) = e^{\tan^{-1}\left(\frac{x}{2}\right)} \left( \frac{2}{4 + x^2} \right)$
$f'(x) = f(x) \left( \frac{2}{4 + x^2} \right)$
$(4 + x^2)f'(x) = 2f(x)$

1. Perform repeated differentiation on the identity from (i) and evaluate the derivatives at $x = 0$.

$f(0) = e^{\tan^{-1}(0)} = 1$

From (a)(i), we are given:
$(4 + x^2)f'(x) = 2f(x)$

Substituting $x = 0$:
$4f'(0) = 2(1) \implies f'(0) = \frac{1}{2}$

Differentiating both sides with respect to $x$:
$(4 + x^2)f''(x) + 2xf'(x) = 2f'(x)$

Substituting $x = 0$:
$4f''(0) + 0 = 2\left(\frac{1}{2}\right) \implies 4f''(0) = 1 \implies f''(0) = \frac{1}{4}$

2. Construct the Maclaurin series for $f(x)$ up to the $x^2$ term.

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \dots$
$f(x) \approx 1 + \frac{1}{2}x + \frac{\frac{1}{4}}{2}x^2$
$f(x) \approx 1 + \frac{1}{2}x + \frac{1}{8}x^2$

3. Determine the value of $x$ required to estimate $e^{\pi/6}$.

We require $f(x) = e^{\pi/6}$:
$e^{\tan^{-1}\left(\frac{x}{2}\right)} = e^{\frac{\pi}{6}}$
$\tan^{-1}\left(\frac{x}{2}\right) = \frac{\pi}{6}$
$\frac{x}{2} = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$
$x = \frac{2}{\sqrt{3}}$

4. Substitute the found value of $x$ into the Maclaurin series to calculate the estimate.

$\text{Estimate} = 1 + \frac{1}{2}\left(\frac{2}{\sqrt{3}}\right) + \frac{1}{8}\left(\frac{2}{\sqrt{3}}\right)^2$
$= 1 + \frac{1}{\sqrt{3}} + \frac{1}{8}\left(\frac{4}{3}\right)$
$= 1 + \frac{\sqrt{3}}{3} + \frac{1}{6}$
$= \frac{7}{6} + \frac{\sqrt{3}}{3} \approx 1.74402$

5. Compute the true value of $e^{\pi/6}$ and calculate the percentage error to show it is approximately 3.3%.

$\text{Actual value} = e^{\frac{\pi}{6}} \approx 1.68809$

$\text{Percentage error} = \frac{|\text{Estimate} - \text{Actual}|}{\text{Actual}} \times 100\%$
$= \frac{|1.74402 - 1.68809|}{1.68809} \times 100\%$
$\approx 3.313\% \approx 3.3\% \text{ (shown)}$

1. Expand the cosine term in the denominator using the trigonometric addition formula.

$g(x) = \frac{1}{\sqrt{2}\left( \cos\frac{x}{a}\cos\frac{\pi}{4} - \sin\frac{x}{a}\sin\frac{\pi}{4} \right)}$
Since $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$, we can simplify the denominator:
$g(x) = \frac{1}{\sqrt{2}\left( \frac{1}{\sqrt{2}}\cos\frac{x}{a} - \frac{1}{\sqrt{2}}\sin\frac{x}{a} \right)} = \frac{1}{\cos\frac{x}{a} - \sin\frac{x}{a}}$

2. Apply small angle approximations to the trigonometric functions.

For sufficiently small $x$, $\cos\frac{x}{a} \approx 1 - \frac{1}{2}\left(\frac{x}{a}\right)^2$ and $\sin\frac{x}{a} \approx \frac{x}{a}$.
$g(x) \approx \frac{1}{1 - \frac{x^2}{2a^2} - \frac{x}{a}}$
$g(x) \approx \left( 1 - \left( \frac{x}{a} + \frac{x^2}{2a^2} \right) \right)^{-1}$

3. Use the standard binomial series expansion up to the term in $x^2$.

Using the standard series $(1-u)^{-1} \approx 1 + u + u^2$ for small $u$, where $u = \frac{x}{a} + \frac{x^2}{2a^2}$:
$g(x) \approx 1 + \left( \frac{x}{a} + \frac{x^2}{2a^2} \right) + \left( \frac{x}{a} + \frac{x^2}{2a^2} \right)^2$

4. Expand and collect terms up to $x^2$ to determine the constants $p$ and $q$.

Expanding up to $x^2$ (ignoring higher powers):
$g(x) \approx 1 + \frac{x}{a} + \frac{x^2}{2a^2} + \left( \frac{x^2}{a^2} \right)$
$g(x) \approx 1 + \frac{1}{a}x + \frac{3}{2a^2}x^2$
Comparing coefficients with $1 + px + qx^2$, we obtain:
$p = \frac{1}{a}, \quad q = \frac{3}{2a^2}$