(c) If \( 3 \tan \beta=-2 \) and \( \beta \in\left(0^{\circ} ; 180^{\circ}\right) \), determine without the use of a calculator and with the aid of a diagram the value of: \( \begin{array}{lll}\text { (1) } \cos \beta & \text { (2) } 2 \sin ^{2} \beta-1 & \text { (3) } \sqrt{13} \cos \beta-13 \sin ^{2} \beta\end{array} \)

First, we start with the equation given, \( 3 \tan \beta = -2 \). This can be rearranged to find \( \tan \beta = -\frac{2}{3} \). Since \( \beta \) is in the range \( (0^{\circ}, 180^{\circ}) \), the tangent is negative in the second quadrant. In the second quadrant, \( \sin \beta \) is positive and \( \cos \beta \) is negative. We can use the definition of tangent, which is \( \tan \beta = \frac{\sin \beta}{\cos \beta} \). Let's denote \( \sin \beta = 2k \) and \( \cos \beta = -3k \) for some positive \( k \): \[ \tan \beta = \frac{2k}{-3k} = -\frac{2}{3}. \] To satisfy the Pythagorean identity \( \sin^2 \beta + \cos^2 \beta = 1 \): \[ (2k)^2 + (-3k)^2 = 1 \\ 4k^2 + 9k^2 = 1 \\ 13k^2 = 1 \\ k^2 = \frac{1}{13} \\ k = \frac{1}{\sqrt{13}}. \] Now we can calculate \( \sin \beta \) and \( \cos \beta \): \[ \sin \beta = 2k = 2 \cdot \frac{1}{\sqrt{13}} = \frac{2}{\sqrt{13}}, \quad \cos \beta = -3k = -3 \cdot \frac{1}{\sqrt{13}} = -\frac{3}{\sqrt{13}}. \] Now let's calculate the required values: (1) **\( \cos \beta \)**: \[ \cos \beta = -\frac{3}{\sqrt{13}}. \] (2) **\( 2 \sin^{2} \beta - 1 \)**: \[ \sin^{2} \beta = \left(\frac{2}{\sqrt{13}}\right)^2 = \frac{4}{13}, \] \[ 2 \sin^{2} \beta - 1 = 2 \cdot \frac{4}{13} - 1 = \frac{8}{13} - 1 = \frac{8}{13} - \frac{13}{13} = -\frac{5}{13}. \] (3) **\( \sqrt{13} \cos \beta - 13 \sin^{2} \beta \)**: \[ \sqrt{13} \cos \beta = \sqrt{13} \cdot \left(-\frac{3}{\sqrt{13}}\right) = -3, \] \[ 13 \sin^{2} \beta = 13 \cdot \frac{4}{13} = 4, \] \[ \sqrt{13} \cos \beta - 13 \sin^{2} \beta = -3 - 4 = -7. \] Summarizing the results: 1. \( \cos \beta = -\frac{3}{\sqrt{13}} \) 2. \( 2 \sin^{2} \beta - 1 = -\frac{5}{13} \) 3. \( \sqrt{13} \cos \beta - 13 \sin^{2} \beta = -7 \)