For number one, gravity is only working in the y-direction. After the ball is kicked, it has an original velocity in both the x- and y- directions, but after that, it is simply acted on by gravity in the y-direction. So, you calculated the initial velocity in the x- and y- directions. Since there are no forces acting on the ball in the x- direction, the equation of the ball's horizontal distance is $x(t) = v_x t = 5t$. In the y-direction, the ball started 11.2m up. Initial velocity is $5\sqrt{3}$. Gravity has an acceleration of $-10$ (I am ignoring stripping units as I prepare to plug this into an equation). So, $y(t) = y_0 + v_y t + a_y t^2 = 11.2 + 5\sqrt{3} t - 10t^2$. You want the value of $t$ when the ball hits the water (sea-level), so you want $y(t)=0$. That is a quadratic equation. Once you know the value of $t$, plug it into $x(t)$ to determine the ball's horizontal position at the time it hit the water.

For 2, you have a force that depends on velocity. In the horizontal direction, the acceleration is $a_x(t) = -0.2v_x(t)$. In the vertical direction, acceleration is $a_y(t) = -10 - 0.2v_y(t)$. That should get you started.