a) The formulas are
vy0 = v0 sin θ
vx0 = v0 cos θ
Plugging in v0 = 40 m/s and θ = 37 deg, we have
vy0 = 24.07 m/s
vx0 = 31.95 m/s
b) When the arrow reaches its highest point, its vertical velocity (vy) is zero. The formula is
vy2 = vy02 + 2ayΔy
The point where vy = 0 corresponds to a specific Δy (the height the arrow goes relative to the archer on top of the castle); vy0 was found in part (a), and ay = -g.
Δy = vy02 / 2g = (24.07 m/s)2 / (2*9.8 m/s2) = 29.56 m
Measured relative to the ground, the arrow reaches a height of 10 m + 29.56 m = 39.56 m.
c) Since the final y position of the arrow is different from its initial position, the process for finding this is a bit more tedious, but still possible. I'll just quote the answer: https://en.wikipedia.org/wiki/Range_of_a_projectile. Using my notation, this is
Δx = (vx0 / g)(vy0 + sqrt[vy02 + 2gy0]
where y0 = 10 m. Plugging in the values gives 169.26 m.