Part a can be solve using the equation of trajectory:
Y = x tana + (g*x^2)/ [2(V0^2)*(cos a)^2]
Where y is the height
X is the length
G is the acceleration due to gravity
Vo Is the initail velocity
a is the angle of trajectory
1.2 = 1.35 tan(0) + (9.81*1.35^2)/ [2(V0^2)*(cos 0)^2]
Solve for V0 = 2.729 m/s
b. can be solve using the formula
v = sqrt(2gy)
= sqrt ( 2*1.2*9.81)
= 4.852 m/s going down ( 0 degree from the horizontal)