Two questions regarding accelerating Atwood's machines I've been trying to solve these two problems for quite some time, but I seem to not be able to grasp the difference between the following similar looking questions.
Problem I 
There is an Atwood's machine facing a constant upwards force of 50 Newtons, at the initial moment two masses connected to the machine are at rest on the ground with masses 2 and 5 kilograms respectively. Find the accelerations of the two masses. The pulley is massless, the string is ideal, gravitational acceleration is $10m/s^2$. 
 
The answer given is $0 m/s^2$ for the 5 kg mass, and $2.5 m/s^2$ for the 2 kg mass.
First of all, my main problem here is that, shouldn't the accelerations of the 2 and 5 kg masses be equal? After all, in a simpler case (where the pulley is hanging from a support), we assume the accelerations of both the masses to be 'a' because they are equal in magnitude, due the relation given by the net work done by tension must be zero: 
$ \Sigma T \cdot x = 0$ 
$ T \cdot x_1 + T \cdot x_2 = 0 $
$ x_1 + x_2 = 0 $ 
$ \frac{d^2 x_1}{dx_1^2} + \frac{d^2 x_2}{dx_2^2} = 0 $ 
$ a_1 + a_2 = 0 => a_1 = -a_2 $ 
Why does the same relation above not apply to the problem? 
Secondly, I tried attempting the problem the following way:
We have a net force of 50N upwards, there are two tensions (both equal, T) acting downwards. As the pulley is massless, net forces on it must sum up to 0. Tension therefore must be 25 N. How can tension be independent of the mass of the two objects? Or is the question simply trying to trick me to believe that there is an external force acting on the pulley while the 50 Newtons is simply a result of something I have missed in the question? 
So tension of 25 N acts upwards on the 2 kg block, while a force of 20 N acts downwards, giving me a total force of 5 N upwards, which gives me an acceleration of $2.5 m/s^2$. For the 5 kg block, I get a tension of 25 N upwards, and 50 N downwards, giving me a net force of -25 N?! Of course, I've made a mistake.
Problem II
An Atwood's machine is facing an upwards force of 80 Newtons, two masses hang from it, which are 2 kilograms and 4 kilograms respectively. What will be the accelerations of the two masses with respect to the pulley? Same diagram as the first, except with newer values.
Now the answer given confuses me further, it's a innocent looking $5 m/s^2$ for both the masses! I get it that the question asks me to find the acceleration with respect to the pulley, but what difference does it make? 
Thank for you reading through this. I've searched across various books and the internet but mechanics is just brutal.
 A: Problem I:
First, as you indicated, the kinematics of the motion has changed relative to the case of a fixed pulley.  So the accelerations of the two masses don't have to be equal in magnitude.  The first step in any pulley problem should be to focus on the kinematics of the motion.  If $y_p$ represents the elevation of the pulley above the table, $y_1$ represents the elevation of mass 1, and $y_2$ represents the elevation of mass 2, then the total length of inextensible string is $$L=(y_p-y_1)+(y_p-y_2)=2y_p-y_1-y_2$$If we differentiate this equation twice with respect to time to get the accelerations, we obtain:$$2a_p-a_1-a_2=0$$
If mass 1 is not accelerating, then $$a_p=a_2/2$$
Second question:  You are imposing a force of 50N manually, and your free-body diagram force balance on the pulley shows that the tension in the string has to be 25 N.  The problem statement is not trying to trick you.  Your own correct analysis has revealed this.  Of course, looking at it another way, the tension actually is related to the masses, since it is this tension that is supporting the weight of mass 2 and also imparting the acceleration to mass 2.  So you are correct to use either interpretation.
Third question:  If you had drawn a free body diagram on mass 1, you would have realized that you have omitted a force from the balance equation on mass 1.  The upward normal force of the table top on mass 1 accounts for the missing 25 N.
Problem 2:
Using 40 N for the tension in the string, from a free body diagram on the 4 kg mass, the net upward force on the 4 kg mass is $40-mg=0$.  This tells us that the 4 kg mass is not accelerating and that the upward normal force exerted by the table on this mass is zero.
From a free body diagram on the 2 kg mass, the net upward force is $40-mg=20 N$.  This tells us that the absolute upward acceleration of the 2 kg mass is $a_2=20/2 = 10\ m/s^2$.  We can now employ the equation we derived under Problem 1 to calculate the absolute upward acceleration of the pulley:$$a_p=\frac{a_2}{2}=5\ m/s^2$$So the upward acceleration of mass 2 relative to the pulley is:  $$a_2-a_p=10-5=5\ m/s^2$$And the upward acceleration of mass 1 relative to the pulley is $$a_1-a_p=0-5=-5\ m/s^2$$.  That is, the pulley is accelerating upward relative to mass 1, or, equivalently, mass 1 is accelerating downward relative to the pulley.
