Let $y = \tan\left(2\tan^{-1}x + \frac{\pi}{4}\right)$.

1. Differentiate $y$ with respect to $x$ using the chain rule.

Given $y = \tan\left(2\tan^{-1}x + \frac{\pi}{4}\right)$.

Differentiating with respect to $x$ using the chain rule:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \sec^2\left(2\tan^{-1}x + \frac{\pi}{4}\right) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\left(2\tan^{-1}x + \frac{\pi}{4}\right)$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \sec^2\left(2\tan^{-1}x + \frac{\pi}{4}\right) \cdot \left( 2 \cdot \frac{1}{1+x^2} \right)$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{1+x^2} \sec^2\left(2\tan^{-1}x + \frac{\pi}{4}\right)$

2. Express the derivative in terms of $y$ using the Pythagorean identity and rearrange.

Using the trigonometric identity $1 + \tan^2\theta = \sec^2\theta$, we substitute $\theta = 2\tan^{-1}x + \frac{\pi}{4}$:
$1 + \tan^2\left(2\tan^{-1}x + \frac{\pi}{4}\right) = \sec^2\left(2\tan^{-1}x + \frac{\pi}{4}\right)$

Since $y = \tan\left(2\tan^{-1}x + \frac{\pi}{4}\right)$, this simplifies to:
$\sec^2\left(2\tan^{-1}x + \frac{\pi}{4}\right) = 1 + y^2$

Substituting this back into our derivative:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{1+x^2}(1+y^2)$

Multiplying both sides by $(1+x^2)$ gives the required result:
$(1+x^2)\frac{\mathrm{d}y}{\mathrm{d}x} = 2(1+y^2)$

1. Evaluate the function $y$ and its first derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $x=0$.

When $x = 0$,
$y = \tan\left(2\tan^{-1}(0) + \frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1$

Using the given equation from (i), $(1+x^2)\frac{\mathrm{d}y}{\mathrm{d}x} = 2(1+y^2)$.
Substituting $x = 0$ and $y = 1$:
$(1+0)\frac{\mathrm{d}y}{\mathrm{d}x} = 2(1+1^2)$
$\frac{\mathrm{d}y}{\mathrm{d}x} = 4$

2. Differentiate the equation from (i) implicitly with respect to $x$.

$\frac{\mathrm{d}}{\mathrm{d}x}\left( (1+x^2)\frac{\mathrm{d}y}{\mathrm{d}x} \right) = \frac{\mathrm{d}}{\mathrm{d}x}\left( 2(1+y^2) \right)$
Using the product rule on the left side and the chain rule on the right side:
$(1+x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2x\frac{\mathrm{d}y}{\mathrm{d}x} = 4y\frac{\mathrm{d}y}{\mathrm{d}x}$

3. Evaluate the second derivative $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ at $x=0$.

Substitute $x = 0$, $y = 1$, and $\frac{\mathrm{d}y}{\mathrm{d}x} = 4$ into the differentiated equation:
$(1+0)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2(0)(4) = 4(1)(4)$
$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 16$

4. Construct the Maclaurin series for $y$ up to the term in $x^2$.

The Maclaurin series expansion is given by $y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots$
$y = 1 + 4x + \frac{16}{2}x^2 + \dots$
$y \approx 1 + 4x + 8x^2$

1. Find the Maclaurin series $\mathrm{g}(x)$ up to the $x^2$ term by using the result from part (i).

From part (i), we have $(1+x^2)\frac{\mathrm{d}y}{\mathrm{d}x} = 2(1+y^2)$.
When $x = 0$, $y = \tan\left(0 + \frac{\pi}{4}\right) = 1$.
Substitute $x=0, y=1$:
$(1+0)\frac{\mathrm{d}y}{\mathrm{d}x} = 2(1+1^2) \implies \frac{\mathrm{d}y}{\mathrm{d}x} = 4$
Differentiating implicitly with respect to $x$:
$(1+x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2x\frac{\mathrm{d}y}{\mathrm{d}x} = 4y\frac{\mathrm{d}y}{\mathrm{d}x}$
Substitute $x=0, y=1, \frac{\mathrm{d}y}{\mathrm{d}x}=4$:
$(1+0)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 0 = 4(1)(4) \implies \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 16$
Thus, the Maclaurin series up to $x^2$ is:
$\mathrm{g}(x) = 1 + 4x + \frac{16}{2!}x^2 = 1 + 4x + 8x^2$

2. Formulate the inequality for $\mathrm{g}(x)$ being within $\pm 0.5$ of $\mathrm{f}(x)$.

The problem requires the difference between the approximation $\mathrm{g}(x)$ and the exact function $\mathrm{f}(x)$ to be at most $0.5$:
$|\mathrm{g}(x) - \mathrm{f}(x)| \le 0.5$
$|1 + 4x + 8x^2 - \tan\left(2\tan^{-1}x + \frac{\pi}{4}\right)| \le 0.5$

3. Solve the inequality graphically using a graphing calculator.

Using a graphing calculator, plot the function $Y_1 = |1 + 4x + 8x^2 - \tan\left(2\tan^{-1}x + \frac{\pi}{4}\right)|$ and the horizontal line $Y_2 = 0.5$.
Find the points of intersection within the specified domain $-0.4 \le x \le 0.4$:
$x \approx -0.35934 \quad \text{and} \quad x \approx 0.22942$
The curve $Y_1$ lies below or on $Y_2$ between these points.
Therefore, the set of values is:
$-0.359 \le x \le 0.229 \quad \text{(to 3 s.f.)}$