What are the accelerations of blocks? I've talked with 2 teachers about this situation:

one teacher said he was completely sure that B have twice the acceleration of A, the other said he was completely sure they have same acceleration. Can you have a better look on it? What do think? Consider it has no friction.
 A: Let the initial length of the bottom segment of rope be $l_1$, the initial length of the middle segment be $l_2$, and the initial length of the top segment be $l_3$.
Since the total length of the rope is constant, we can write
$$l_1 + l_2 + l_3 = K$$
Now, displace the block $A$ by $\Delta x_A$ down the slope and then
$$(l_1 + \Delta x_A)  + (l_2 + \Delta x_A) + (l_3 + \Delta l_3) = K$$
Thus
$$\Delta l_3 = -2 \Delta x_A$$
So, the length of the segment $l_3$ decreases twice as much as the displacement of block $A$.
However, this isn't the entire story.  The top-most pulley moves down the slope with block $A$ and so the displacement of block $B$ is 
$$\Delta x_B = \Delta l_3 + \Delta x_A = -\Delta x_A$$
And so we conclude that the blocks have accelerations of equal magnitude and opposite direction.
A: The two blocks have the same acceleration:

The position of block B is determined by $l_1$ and $l_3$, that of A is determined by $l_3$ or $l_4$ (as they vary by the same amount).
Let $l_1$ increase by a distance $d$, then $(l_3+l_4)$ will decrease by the same $d$ (in case the string is inextensible and $l_2$ remains constant), and they will share this decrease equally, each one will decrease a $\dfrac{d}{2}$. Therefore, the net motion of B will be $d-\dfrac{d}{2}=\dfrac{d}{2}$ downwards, while that of A will be $\dfrac{d}{2}$ upwards. You can derive w.r.t time to get the accelerations.
A: 
$$l=x_A+(x_A-k_1)+k_2+k_3+2\pi R$$
$$k_3=x_B-2R-(x_A-k_1)$$
$$l=x_A+x_B+constant$$
so, we have:
$$0=v_A+v_B$$
and$$a_A=-a_B$$
A: The accelerations are the same. They can't be different, since otherwise $B$ would move faster than $A$ and the rope wouldn't be tight anymore.
The reason your teacher thought that they are different is a common mistake: under the assumption $F_{AB} = -F_{BA}$ and $F=ma$ (force on A should be force from B and vice versa) you might think that same $F$ and different $m$ should result in different $a$, e.g. $a_{A}=\frac{F_{AB}}{m_{A}}=\frac{m_{A}-m_{B}}{m_{A}}g^{*}$ and $a_{B}=\frac{F_{BA}}{m_{B}}=-\frac{m_{A}-m_{B}}{m_{B}}g^{*}$ (where $g^{*} = g * sin(25°)$ in your example).
The reason that's wrong is the following: the resulting force, that is what's left if you substract the gravitational forces on both masses, $F=({m_{A}-m_{B}}) g^{*}$, has to accelerate BOTH masses. Image how the blocks would move without gravity, they are just two masses bound together behind each other that move in the same direction (= "the same direction as the rope"). Gravitational forces on $B$ are already "taken care of" by some part of the gravitional force on $A$ via the pulley. So if you (or the earth) pull on $A$ you have to move both $A$ and $B$. 
So you get $a = \frac{F}{m_{A}+m_{B}} = \frac{{m_{A}-m_{B}}}{m_{A}+m_{B}}g^{*}$ (for both blocks).
