(b) If \( \tan \theta=\frac{5}{12} \) and \( \sin \theta>0 \), determine by means of a diagram: \( \begin{array}{ll}\text { (1) } 13 \cos \theta & \text { (2) } \cos ^{2} \theta+\sin ^{2} \theta \end{array} \)

To solve for \( 13 \cos \theta \) and \( \cos^2 \theta + \sin^2 \theta \) given that \( \tan \theta = \frac{5}{12} \) and \( \sin \theta > 0 \), we can start by interpreting the tangent ratio in a right triangle where the opposite side is 5 and the adjacent side is 12. Using the Pythagorean theorem, the hypotenuse \( h \) can be calculated as follows: \[ h = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13. \] Now, we can find \( \sin \theta \) and \( \cos \theta \): \[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}, \quad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}. \] Now we can calculate the requested expressions: 1) \[ 13 \cos \theta = 13 \cdot \frac{12}{13} = 12. \] 2) Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \): \[ \cos^2 \theta + \sin^2 \theta = \left(\frac{12}{13}\right)^2 + \left(\frac{5}{13}\right)^2 = \frac{144}{169} + \frac{25}{169} = \frac{169}{169} = 1. \] Thus, the final answers are: 1) \( 13 \cos \theta = 12 \) 2) \( \cos^2 \theta + \sin^2 \theta = 1 \)