Answer
(a) The maximum range is $\frac{v_i^2}{g}$
(b) The maximum range occurs at a launch angle of $45^{\circ}$
Work Step by Step
$R = \frac{2~v_i^2~sin~\theta~cos~\theta}{g} = \frac{v_i^2~sin~2\theta}{g}$
(a) The maximum range occurs when $sin~2\theta = 1$. Then the maximum range is $\frac{v_i^2}{g}$
(b) The maximum range occurs when $sin~2\theta = 1$. We can find the angle $\theta$:
$sin~2\theta = 1$
$2\theta = sin^{-1}~(1)$
$2\theta = 90^{\circ}$
$\theta = 45^{\circ}$
The maximum range occurs at a launch angle of $45^{\circ}$
