Answer: 
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m 
and f =  v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69 
given that A = 2.20 mm = 2.2×10⁻³
so  = A × ω
 = A × ω
 = 2.2×10⁻³ × 2722.69 m/s
 = 2.2×10⁻³ × 2722.69 m/s 
 =  5.9899 m/s
 =  5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin(  0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
 = A' × ω
 = A' × ω
 = 1.555×10⁻³ × 2722.69
 = 1.555×10⁻³ × 2722.69 
 = 4.2338 m/s
 = 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s