Answer
(a) $a = 3.1 ~m/s^2$
(b) $F_T = 2.0~N$
Work Step by Step
(a) Let's consider the system of both blocks. We can use a force equation to find the acceleration. Let $m_A$ and $m_B$ be $m$ since both blocks have the same mass.
$(2m)a = \sum F$
$2ma = 2mg ~sin(\theta) - mg ~cos(\theta)\cdot \mu_A - mg ~cos(\theta)\cdot \mu_B$
$2a = 2g ~sin(\theta) - g ~cos(\theta)\cdot (0.20) - g ~cos(\theta)\cdot (0.30)$
$2a = 2g ~sin(\theta) - g ~cos(\theta)\cdot (0.50)$
$2a = 2g ~sin(\theta) - 2g ~cos(\theta)\cdot (0.25)$
$a = g ~sin(\theta) - g ~cos(\theta)\cdot (0.25)$
$a = (9.80 ~m/s^2) ~sin(32^{\circ}) - (9.80 ~m/s^2) ~cos(32^{\circ}) \cdot (0.25)$
$a = 3.1 ~m/s^2$
(b) Let's consider the system of block B. We can use a force equation to find the tension $F_T$ in the cord:
$\sum F = ma$
$F_T + mg ~sin(\theta) - mg ~cos(\theta) \cdot \mu_B = ma$
$F_T = ma - mg ~sin(\theta) + mg ~cos(\theta) \cdot \mu_B$
$F_T = (5.0 ~kg)(3.1 ~m/s^2) - (5.0 ~kg)(9.80 ~m/s^2) ~sin(32^{\circ}) + (5.0 ~kg)(9.80 ~m/s^2)~cos(32^{\circ})\cdot (0.30)$
$F_T = 2.0~N$
