Determine the accelerations of the blocks A and B for the three...

Let $T$ be the tension in the string for each case.

For situation (a):
Let the downward acceleration of block $A$ be $a$. Since the string is continuous, if block $A$ moves down by a distance $x$, the movable pulley (and block $B$) must move up by a distance $x/2$. Thus, the upward acceleration of block $B$ is $a/2$.
The equations of motion are:
For block $A$ ($4\text{ kg}$): $4g - T = 4a$
For block $B$ ($5\text{ kg}$): $2T - 5g = 5$\left(\dfrac{a}{2}\right)
Multiplying the first equation by $2$ gives $8g - 2T = 8a$.
Adding this to the second equation:
$8g - 5g = 8a +$\dfrac{5a}{2}
$3g =$\dfrac{21a}{2} \Rightarrow$a =$\dfrac{6g}{21} = \dfrac{2g}{7}
Thus, the acceleration of $A$ is \dfrac{2g}{7} downward, and the acceleration of $B$ is \dfrac{g}{7} upward.

For situation (b):
Let the rightward acceleration of block $A$ be $a$. If block $A$ moves right by $x$, the movable pulley moves down by $x/2$. Thus, the downward acceleration of block $B$ is $a/2$.
The equations of motion are:
For block $A$ ($2\text{ kg}$): $T = 2a$
For block $B$ ($5\text{ kg}$): $5g - 2T = 5 \left(\dfrac{a}{2}\right)$
Substituting $T = 2a$ into the second equation:
$5g - 2(2a) = \dfrac{5a}{2}$
$5g - 4a = \dfrac{5a}{2} \Rightarrow 5g = \dfrac{13a}{2} \Rightarrow a = \dfrac{10g}{13}$
Thus, the acceleration of $A$ is $\dfrac{10g}{13}$ right, and the acceleration of $B$ is $\dfrac{5g}{13}$ downward.

For situation (c):
Let the downward acceleration of block $A$ be $a$. If block $A$ moves down by $x$, the movable pulley moves up by $x/2$. Thus, the upward acceleration of block $B$ is $a/2$.
The equations of motion are:
For block $A$ ($2\text{ kg}$): $2g - T = 2a$
For block $B$ ($1\text{ kg}$): $2T - 1g = 1$\left(\dfrac{a}{2}\right)
Multiplying the first equation by $2$ gives $4g - 2T = 4a$.
Adding this to the second equation:
$4g - g = 4a +$\dfrac{a}{2}
$3g =$\dfrac{9a}{2} \Rightarrow$a =$\dfrac{6g}{9} = \dfrac{2g}{3}
Thus, the acceleration of $A$ is \dfrac{2g}{3} downward, and the acceleration of $B is \dfrac{g}{3} upward.$