Answer
$F_T = 3.16~N$
Work Step by Step
We can consider the the forces on $m_1$ to find an expression for $a$:
$m_1~g~sin~\theta+F_T-m_1~g~\mu_1~cos~\theta = m_1~a$
$a = \frac{m_1~g~sin~\theta+F_T-m_1~g~\mu_1~cos~\theta}{m_1}$
We can consider the the forces on $m_2$ to find an expression for $a$:
$m_2~g~sin~\theta+F_T-m_2~g~\mu_2~cos~\theta = m_2~a$
$a = \frac{m_2~g~sin~\theta+F_T-m_2~g~\mu_2~cos~\theta}{m_2}$
We can equate the two expressions to find the tension $F_T$:
$a = \frac{m_1~g~sin~\theta+F_T-m_1~g~\mu_1~cos~\theta}{m_1} = \frac{m_2~g~sin~\theta+F_T-m_2~g~\mu_2~cos~\theta}{m_2}$
$m_1~m_2~g~sin~\theta+F_T~m_2-m_1~m_2~g~\mu_1~cos~\theta = m_1~m_2~g~sin~\theta+F_T~m_1-m_1~m_2~g~\mu_2~cos~\theta$
$F_T~(m_2-m_1) = m_1~m_2~g~\mu_1~cos~\theta -m_1~m_2~g~\mu_2~cos~\theta$
$F_T = \frac{m_1~m_2~g~\mu_1~cos~\theta -m_1~m_2~g~\mu_2~cos~\theta}{m_2-m_1}$
$F_T = \frac{(1.65~kg)~(3.30~kg)~(9.8~m/s^2)~(0.226)~(cos~30.0^{\circ}) - (1.65~kg)~(3.30~kg)~(9.8~m/s^2)~(0.113)~(cos~30.0^{\circ})}{3.30~kg-1.65~kg}$
$F_T = 3.16~N$
