b. $\displaystyle tanu = -\frac{8}{5}, restriction: \frac{3\pi}2 < u < 2\pi$

tangent is "opposite side over near side" so imagine a triangle with "opposite side" of length 8 and "near side" of length 5. You can use the Pythagorean theorem to get the length of the hypotenuse and so sin(u) and cos(u). Since u is in the fourth quadrant, sine is negative and cosine is positive. Since u is in the fourth quadrant, u/2 is in the second quadrant, where cosine is negative.

Please solve it step by step so I can understand it too. xD