Answer
(a) The initial velocity is directed at $44.5^\circ$ above the horizontal.
(b) The initial speed is $30\text{ m/s}$.
Work Step by Step
(a) The horizontal distance traveled is given by the product of the horizontal component of the velocity and the time of the flight
$$D=v_0\cos\theta\cdot t_f$$
The time of the flight is given by
$$t_f=\frac{2v_0\sin\theta}{g}.$$
Dividing previous two equations we obtain
$$\frac{D}{t_f}=\frac{v_0t_f\cos\theta}{\frac{2v_0\sin\theta}{g}}=\frac{gt_f\cos\theta}{2\sin\theta}=\frac{gt_f}{2\tan\theta}.$$
This further gives
$$\tan\theta=\frac{gt_f^2}{2D}\Rightarrow\theta=\arctan\frac{gt_f^2}{2D}=\arctan\frac{9.81\text{ m/s}^2\cdot(4.30\text{ s})^2}{2\cdot92.2\text{ m}}=44.5^\circ,$$
so the initial velocity is directed $44.5^\circ$ above the horizontal.
(b) From the equation for $D$ we get
$$v_0=\frac{D}{t_f\cos\theta}=\frac{92.2\text{ m}}{4.30\text{ s }\cdot\cos44.5^\circ}=30\text{ m/s}.$$
