Now then, here's how you get the first answer:
The bag's original velocity (yes, I know the problem is asking for speed, but we'll get to that in a minute) is -1.50 m/s, since it's moving downward with the helicopter. When it's released, it undergoes a downward acceleration of -9.81 m/s² due to gravity (acceleration is also a vector). We need to find its velocity after 2.00 seconds. There's a formula for that:
a = (vf - vo) / t
Where a is acceleration in m/s², vo is original velocity and vf is final velocity (both in m/s), and t is time in s. Remember to use your positive and negative signs since we're dealing with directions.
-9.81 m/s² = (vf - (-1.50 m/s)) / 2.00 s
Now solve for vf:
-19.62 m/s = vf - (-1.50 m/s)
vf = -21.1 m/s
That's the bag's VELOCITY after 2.00 s, but not its speed. Speed has no direction, so the speed should be reported as just 21.1 m/s.
(b) We need to consider two motions: the helicopter's constant downward movement of -1.50 m/s, and the bag's downward acceleration at -9.81 m/s². Let's start with the helicopter and figure out how far it will be below the "drop height" after 2.00 s. Remember, its velocity is constant.
v = Δx / t
Where v = velocity, Δx = displacement (another vector), and t = time
-1.50 m/s = Δx / 2.00 s
Δx = -3.00 m
The helicopter will be 3.00 m below the "drop height" 2.00 s after the drop. Now for the bag. Since the bag is accelerating downward we have to use a different formula to calculate its displacement:
Δx = vo*t + 1/2at²
Δx = (-1.50 m/s)(2.00 s) + (1/2)(-9.81 m/s²)(2.00 s)²
Δx = (-3.00 m) + (-4.91 m/s²)(4.00 s²)
Δx = -3.00 m - 19.6 m
Δx = -22.6 m
So the helicopter is 3.00 meters BELOW the drop height and the bag is 22.6 meters BELOW it. The displacement between them is y2 - y1, or (-22.6 m - (-3.00 m)) = -19.6 m. But displacement is a vector, and you want distance, which is a scalar quantity. No problem, just drop the negative sign and you've got the distance: 19.6 m.