Pet Peeve Warning!! Danger! Danger!
What the heck happened to the all the units in your answers??
(Pet peeve not over, to be continued indefinitely.)
I'll start this from the beginning, since it looks like you are applying final formulas to get your answer.
I have a diagram that puts the 2 kg mass on the left side of the pulley and the 8 kg mass on the right. To analyze this note that the pulley is massless and frictionless and we have an ideal string, so neither really influences the problem except that the pulley changes the direction of the tension in the string. (And the tension along an ideal string is the same at all points.)
So. Free-Body Diagram on the 2 kg mass (m1). We have a weight (w1) acting downward and a tension (T) acting upward. Since we know that the other mass is larger than this one we can state that the acceleration of this mass is upward. So I am going to go with my usual convention and put the +y direction in upward: in the direction of the acceleration of the object. So Newton's 2nd says:
For the FBD on the 8 kg mass (m2), we have a weight (w2) acting downward and a tension (T) acting upward. Since the two masses are connected by an ideal string both tensions will be equal and so will the accelerations. BUT the acceleration on this mass is now downward. To be consistent with the first FBD I am going to define +y to be downward here, else we have to play with an extra negative sign. So Newton's 2nd says:
<-- Note the effect of the change in positive direction here.
So we have a system of equations for a and T:
We may solve these in any way we wish, substitution being the method most likely to be used by a Physicist, and obtain:
(The appropriate directions of these are given by the FBDs.)
So I get that the magnitude of the acceleration is:
The magnitude of the tension is:
(To the correct number of sig. digs.)
Now, we know the acceleration of the 2.00 kg mass is upward and constant, and the initial velocity of it is 2.80 m/s downward. In the coordinate system of the FBD for the 2.00 kg mass (+y is upward), thus we use a = 5.88 m/s^2 upward (ie. use a as a positive number) and v0 = 2.80 m/s downward (ie. use v0 as a negative number.)
So we want the lowest distance this mass goes. At this point v = 0 m/s. Thus:
v^2 = v_0^2 + 2a(y - y_0) <-- LaTeX won't do this for some reason.
where y represents the lowest position of the mass and v = 0 m/s. I will set the 0 point of the coordinate system to be where the mass starts at t = 0 s.
So this is a very long message stating your answers are correct, except for ignoring the units on your answers.