Answer
(a) The initial direction is $32^\circ$ above the horizontal.
(b) The time of the flight is $t_f=0.76\text{ s}$.
Work Step by Step
(a) The distance $D$ traveled is equal to the horizontal component of the velocity (which is constant during the motion) multiplied by the time of flight i.e.
$$t_f=\frac{2v_0\sin\theta}{g}\Rightarrow D=v_xt_f=\frac{2v_0^2\sin\theta\cos\theta}{g}.$$
We can use the trigonometric identity that
$$\sin2\theta=2\sin\theta\cos\theta$$
so the expression for distance becomes
$$D=\frac{v_0^2}{g}\sin2\theta.$$
Now we find
$$\sin2\theta=\frac{gD}{v_0^2}\Rightarrow2\theta=\arcsin\frac{gD}{v_0^2}\Rightarrow\theta=\frac{1}{2}\arcsin\frac{gD}{v_0^2},$$
and finally
$$\theta=\frac{1}{2}\arcsin\frac{gD}{v_0^2}=\frac{1}{2}\arcsin\frac{9.81\text{ m/s}^2\cdot4.6\text{ m}}{7.1^2}=32^\circ$$
(b) The time of the flght is now readilly found:
$$t_f=\frac{2v_0\sin\theta}{g}=\frac{2\cdot7.1\text{ m/s }\cdot\sin32^\circ}{9.81\text{ m/s}^2}=0.76\text{ s}.$$
