Given that theta = (5)/(13), and theta is acute, find the value of ( theta + theta)/( theta).

Since $\theta$ is acute, we use the identity $\sin^2 \theta + \cos^2 \theta = 1$.
$$
\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{169 - 25}{169} = \frac{144}{169}
$$
Since $\theta$ is acute, $\cos \theta > 0$.
$$
\cos \theta = \sqrt{\frac{144}{169}} = \frac{12}{13}
$$
Now, find $\tan \theta$:
$$
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{5/13}{12/13} = \frac{5}{12}
$$

$$
\frac{\frac{5}{12} + \frac{12}{13}}{\frac{5}{13}}
$$

First, simplify the numerator:
$$
\frac{5}{12} + \frac{12}{13} = \frac{5 \times 13 + 12 \times 12}{12 \times 13} = \frac{65 + 144}{156} = \frac{209}{156}
$$
Now, divide by $\sin \theta$:
$$
\frac{\frac{209}{156}}{\frac{5}{13}} = \frac{209}{156} \times \frac{13}{5}
$$
Since $156 = 12 \times 13$:
$$
\frac{209}{12 \times 13} \times \frac{13}{5} = \frac{209}{12 \times 5} = \frac{209}{60}
$$
$$
\boxed{\frac{209}{60}}
$$

$$
\sin 5x = \sin (90^\circ - 4x)
$$

$$
5x = 90^\circ - 4x
$$

$$
5x + 4x = 90^\circ
$$
$$
9x = 90^\circ
$$
$$
x = \frac{90^\circ}{9}
$$
$$
x = 10^\circ
$$
$$
\boxed{10^\circ}
$$

$$
\cos (30^\circ + \theta) = \sin (90^\circ - (30^\circ + \theta))
$$
$$
\cos (30^\circ + \theta) = \sin (90^\circ - 30^\circ - \theta)
$$
$$
\cos (30^\circ + \theta) = \sin (60^\circ - \theta)
$$

$$
\sin (20^\circ + \theta) = \sin (60^\circ - \theta)
$$
$$
20^\circ + \theta = 60^\circ - \theta
$$

$$
\theta + \theta = 60^\circ - 20^\circ
$$
$$
2\theta = 40^\circ
$$
$$
\theta = \frac{40^\circ}{2}
$$
$$
\theta = 20^\circ
$$
$$
\boxed{20^\circ}
$$

Since $\theta$ is acute, we use the identity $\sin^2 \theta + \cos^2 \theta = 1$.
$$
\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}
$$
Since $\theta$ is acute, $\cos \theta > 0$.
$$
\cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}
$$
Now, find $\sec \theta$:
$$
\sec \theta = \frac{1}{\cos \theta} = \frac{1}{4/5} = \frac{5}{4}
$$
Next, find $\tan \theta$:
$$
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}
$$
Then, find $\cot \theta$:
$$
\cot \theta = \frac{1}{\tan \theta} = \frac{1}{3/4} = \frac{4}{3}
$$

$$
\frac{5}{4} + \frac{4}{3}
$$

$$
\frac{5}{4} + \frac{4}{3} = \frac{5 \times 3 + 4 \times 4}{4 \times 3} = \frac{15 + 16}{12} = \frac{31}{12}
$$
$$
\boxed{\frac{31}{12}}
$$