The trajectory of the ball is an horizontal circle of radius R = L sin(A) where L (= 1.7 m) is the length of the string and A is the angle (called "theta" in the picture) which it makes with the vertical.
If the ball travels at speed v, it undergoes a centripetal (horizontal) acceleration v^2/R (see first link below for the general theorem--where R is the radius of curvature of the trajectory-- which says that the above formula always give the component of the acceleration which is perpendicular to the trajectory, even when the speed is not constant and/or when the trajectory is not a circle). This is the basis for the following answers to your questions:
(a) If g is the acceleration of gravity then v^2/R = g tan(A).
So, v^2 is Rg tan(A) = gL sin(A) tan(A)
With A = 30° and g = 9.80665 m/s^2 (see second link below) a length L=1.7 m corresponds to v = 2.19376 m/s
(b) The formula is v^2 = gL sin(A) tan(A) = gL (1-x^2)/x where x is cos(A). This is a quadratic equation in x, namely:
x^2 + (v^2/gL) x -1 = 0
Letting v be 4 m/s and solving for a positive value of x below 1, we obtain x = 0.6293 and, therefore, A = 51° (almost exactly).
(c) The tension of the rope is mg / cos(A) = mg / x.
When m is 0.5 kg, this is equal to 9.1 N when x is about .538827 (incidentally, this corresponds to an angle of about 57.4°, pretty close to 1 radian). Using the aforementioned relation v^2 = gL (1-x^2)/x, we obtain the value of v at which the rope would break, namely: v = 4.6858 m/s.