Given theta = (1)/(4).

Here are the solutions for the given trigonometric expressions:

a) Find $\sin \theta$ Step 1: Use the identity $\sin^2 \theta + \cos^2 \theta = 1$. Given $\cos \theta = \frac{1}{4}$. $\sin^2 \theta + \left(\frac{1}{4}\right)^2 = 1$ $\sin^2 \theta + \frac{1}{16} = 1$ $\sin^2 \theta = 1 - \frac{1}{16}$ $\sin^2 \theta = \frac{16}{16} - \frac{1}{16}$ $\sin^2 \theta = \frac{15}{16}$ Since $\theta$ is acute, $\sin \theta$ is positive. $\sin \theta = \sqrt{\frac{15}{16}}$ \sin \theta = \frac{\sqrt{15}{4}}

b) Find $\tan^2 \theta$ Step 2: Use the identity $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$. From Step 1, we have $\sin^2 \theta = \frac{15}{16}$. We are given $\cos \theta = \frac{1}{4}$, so $\cos^2 \theta = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$. $\tan^2 \theta = \frac{\frac{15}{16}}{\frac{1}{16}}$ $\tan^2 \theta = \frac{15}{16} \times \frac{16}{1}$ $\tan^2 \theta = 15$

c) Find $\frac{1}{\sin^2 \theta}$ Step 3: Use the value of $\sin^2 \theta$ from Step 1. We found $\sin^2 \theta = \frac{15}{16}$. $\frac{1}{\sin^2 \theta} = \frac{1}{\frac{15}{16}}$ \frac{1}{\sin^2 \theta} = \frac{16{15}}