Your equations are correct, though personally I would have chosen a different set of letters. For instance I would have called all the accelerations using the letter a: A, ap, a1, a2. But as long as they are labeled clearly on your diagram you are fine. In the future, though, it would be nice to have a larger and better labeled picture. Remember if your teacher can't read it they are not likely to give you full value even for a correct answer.
The good news is that I disagree with your text as well. I'm getting a = (2/5)g.
-Dan
Okay, let's do this the way I wanted to originally. All we need are FBDs on the 8 kg mass and the total weight acting on the pulley.
8 kg mass: +x up the slope, +y in the direction of the normal force.
$\displaystyle \sum F_x = t - f_k - w_x = 8 a$
etc.
$\displaystyle t - 4 g - 4 g = 8 a$
Pulley: +x downward.
The mass of the "pulley" is 2 g + 3 g + 5 g.
$\displaystyle \sum F_x = -t + 10 g = 10 a$
Two equations, two unknowns. I get a = g/9 and t = 80g/9
Clearly the book is wrong but there must also something wrong in the equations originally given by markosheehan, though I didn't find anything.
-Dan
Rechecked my work and arrived at the book solution of $a=\dfrac{3g}{35}$
Since the lower pulley system is accelerating downward at $a$, the apparent weights of the 5kg and 3kg mass are $5(g-a)$ and $3(g-a)$ respectively. Let the tension in the lower pulley system string be $T_2$ and the acceleration of the 5kg and 3kg masses within their reference frame be $a'$.
$5(g-a)-T_2=5a'$
$T_2-3(g-a)=3a'$
$a'=\dfrac{g-a}{4} \implies T_2=\dfrac{15(g-a)}{4}$
Let the tension in the string attached to the sliding 8kg mass be $T_1$
Force equations for the 8kg mass and 2kg pulley are ...
$T_1 - 8g[\mu \cos(30) + \sin (30)] = 8a \implies T_1 - 8g = 8a$
$2T_2 + 2g - T_1 = 2a \implies \dfrac{15(g-a)}{2} + 2g - T_1 = 2a \implies \dfrac{19g}{2} -T_1 = \dfrac{19a}{2}$
adding the last two equations yields ...
$\dfrac{3g}{2}=\dfrac{35a}{2} \implies a= \dfrac{3g}{35}$
$5(g-a)-T_2=5a'$
$T_2-3(g-a)=3a'$
do the equations for the lower pulley 5g-t2=5(a+b) and t2-3g=(b-a) work?. in your answer why dont you account for the acceleration of both pulleys you wrote ^. should it the acceleration in the equation not be 5(a'+a) and 3(a'-a)
rearranging the two scalar equations I posted ...do the equations for the lower pulley 5g-t2=5(a+b) and t2-3g=(b-a) work?
$5(g-a) - T_2 = 5a' \implies 5g - 5a - T_2 = 5a' \implies 5g - T_2 = 5(a+a')$
$T_2 - 3(g-a) = 3a' \implies T_2 - 3g + 3a = 3a' \implies T_2 - 3g = 3(a'-a)$
see if that translates to the variables you chose to use ...