It is given that $\mathrm{f}(x) = \begin{cases} \tan\left(\frac{\pi}{4a} x\right) & \text{for } -a < x \le a, \\ \sin\left(\frac{\pi}{2a} x\right) & \text{for } a < x \le 3a, \end{cases}$ and that $\mathrm{f}(x) = \mathrm{f}(x-4a)$ for all real values of $x$, where $a$ is a positive real constant.

1. Identify the key points and shapes of the given pieces for one base period.

Consider the base period from $x = -a$ to $x = 3a$.
For $-a < x \le a$:
$y = \tan\left(\frac{\pi}{4a} x\right)$

• As $x \to -a^+$, $y \to -1$.
• At $x = 0$, $y = 0$.
• At $x = a$, $y = 1$.

For $a < x \le 3a$:
$y = \sin\left(\frac{\pi}{2a} x\right)$

• At $x = a$, $y = \sin\left(\frac{\pi}{2}\right) = 1$ (joining the tangent piece continuously).
• At $x = 2a$, $y = \sin(\pi) = 0$.
• At $x = 3a$, $y = \sin\left(\frac{3\pi}{2}\right) = -1$.

Thus, one complete cycle has a length of $3a - (-a) = 4a$, consisting of an increasing segment from $(-a, -1)$ to $(a, 1)$ crossing at $(0, 0)$, and a decreasing segment from $(a, 1)$ to $(3a, -1)$ crossing at $(2a, 0)$.

2. Apply the periodicity to extend the key features over the full required domain.

Using the periodic property $\mathrm{f}(x) = \mathrm{f}(x-4a)$, the function repeats every $4a$. We translate the key features across the domain $-4a \le x \le 7a$:

• $x$-intercepts ($y=0$): Add and subtract multiples of $4a$ to $\{0, 2a\}$ to obtain $-4a, -2a, 0, 2a, 4a, 6a$.
• Maximum points ($y=1$): Translate the peak at $x=a$ by steps of $4a$ to obtain $-3a, a, 5a$.
• Minimum points ($y=-1$): Translate the trough at $x=3a$ by steps of $4a$ to obtain $-a, 3a, 7a$.

3. Describe the assembly of the final sketch on the coordinate axes.

Draw the full sketch for $-4a \le x \le 7a$:

1. Label the $x$-axis from $-4a$ up to $7a$, and the $y$-axis with $1$ and $-1$.
2. Plot the maximums, minimums, and $x$-intercepts identified in the previous step.
3. Connect the points with the appropriate shapes: from each minimum, draw an increasing curve to the next maximum (concave down then concave up), followed by a decreasing sine curve to the next minimum.
4. Ensure the graph explicitly starts at the root $(-4a, 0)$ and terminates at the minimum $(7a, -1)$.

1. Identify the symmetry of the function from the graph to deduce the relationship

From the sketch in part (a), the graph of $y = \mathrm{f}(x)$ has rotational symmetry of order 2 about the origin $(0,0)$. This geometric property indicates that $\mathrm{f}(x)$ is an odd function.

Therefore, we can write down the relationship:
$\mathrm{f}(-x) = -\mathrm{f}(x)$

1. Use the odd function property from part (b) to simplify the integral over the symmetric interval $[-2a, 2a]$.

From part (b), $\mathrm{f}(x)$ is an odd function ($\mathrm{f}(-x) = -\mathrm{f}(x)$), so its integral over any symmetric interval is zero:
$\int_{-2a}^{2a} \mathrm{f}(x) \,\mathrm{d}x = 0$
Thus, we can split the required integral:
$\int_{-2a}^{4a} \mathrm{f}(x) \,\mathrm{d}x = \int_{-2a}^{2a} \mathrm{f}(x) \,\mathrm{d}x + \int_{2a}^{4a} \mathrm{f}(x) \,\mathrm{d}x = 0 + \int_{2a}^{4a} \mathrm{f}(x) \,\mathrm{d}x$

2. Apply the periodic property of $\mathrm{f}(x)$ to shift the integration bounds into intervals where the function is explicitly defined.

Using the periodicity $\mathrm{f}(x) = \mathrm{f}(x-4a)$, we can shift the interval $[2a, 4a]$ by $-4a$:
$\int_{2a}^{4a} \mathrm{f}(x) \,\mathrm{d}x = \int_{-2a}^{0} \mathrm{f}(x) \,\mathrm{d}x = \int_{-2a}^{-a} \mathrm{f}(x) \,\mathrm{d}x + \int_{-a}^{0} \mathrm{f}(x) \,\mathrm{d}x$
For the first term, we apply $\mathrm{f}(x) = \mathrm{f}(x+4a)$ to shift the interval to $[2a, 3a]$, where the sine definition applies:
$\int_{-2a}^{-a} \mathrm{f}(x) \,\mathrm{d}x = \int_{2a}^{3a} \mathrm{f}(x) \,\mathrm{d}x = \int_{2a}^{3a} \sin\left(\frac{\pi}{2a}x\right) \,\mathrm{d}x$

3. Evaluate the definite integral involving the sine function over $[2a, 3a]$.

$\int_{2a}^{3a} \sin\left(\frac{\pi}{2a}x\right) \,\mathrm{d}x = \left[ -\frac{2a}{\pi} \cos\left(\frac{\pi}{2a}x\right) \right]_{2a}^{3a}$
$= -\frac{2a}{\pi} \left( \cos\frac{3\pi}{2} - \cos\pi \right)$
$= -\frac{2a}{\pi} (0 - (-1)) = -\frac{2a}{\pi}$

4. Evaluate the definite integral involving the tangent function over $[-a, 0]$.

Using the definition $\mathrm{f}(x) = \tan\left(\frac{\pi}{4a}x\right)$ for $x \in (-a, 0]$:
$\int_{-a}^{0} \tan\left(\frac{\pi}{4a}x\right) \,\mathrm{d}x = \left[ \frac{4a}{\pi} \ln\left|\sec\left(\frac{\pi}{4a}x\right)\right| \right]_{-a}^0$
$= \frac{4a}{\pi} \left( \ln(\sec 0) - \ln\left|\sec\left(-\frac{\pi}{4}\right)\right| \right)$
$= \frac{4a}{\pi} \left( \ln 1 - \ln\sqrt{2} \right)$
$= \frac{4a}{\pi} \left( 0 - \frac{1}{2}\ln 2 \right) = -\frac{2a}{\pi}\ln 2$

5. Combine the evaluated parts to deduce the final values of $k$ and $m$.

Adding the integrals together:
$\int_{-2a}^{4a} \mathrm{f}(x) \,\mathrm{d}x = -\frac{2a}{\pi} - \frac{2a}{\pi}\ln 2 = -\frac{2a}{\pi}(1 + \ln 2)$
Comparing this result to the given form $\frac{ka}{\pi}(1 + \ln m)$, we determine that:
$k = -2$ and $m = 2$.