Sled A and sled B, whose masses are 40.0kg and 80.0kg respectively, are attached by a massless bar. The sleds slide down a snow-covered hill

You write down all the forces on each mass, then resolve them parallel and perpendicular to the surface of the hill.  The perpendicular forces on each mass must balance; the net force down the hill must give rise to an acceleration (the same for both masses since they are rigidly fixed together).  

You don't say which mass is lower so I will assume mass A is lower (this will only affect the sign of the tension);

 

Mass A:

Perpendicular:   F_NA = m_Ag*cos(θ)                    ...(1)

where F_NA is normal force, m_A is mass of A;  g is gravitational acceleration;  θ is angle to horizontal.

Parallel:           m_A*a = m_Ag*sin(θ) - μ_AF_NA - T  ...(2)

where a is acceleration; μ_A is coefficient of friction for mass A; T is tension in the bar.

 

Mass B:

Perpendicular:  F_NB = m_Bg*cos(θ)   ...(3)

Parallel:           m_B*a = m_Bg*sin(θ) - μ_BF_NB + T  ...(4)

 

Use (1) in (2) and (3) in (4)

m_A*a = m_Ag*sin(θ) - μ_Am_Ag*cos(θ) - T     ...(5)

m_B*a = m_Bg*sin(θ) - μ_Bm_Bg*cos(θ)  + T   ...(6)

 

Add equations (5) and (6):

(m_A + m_B)*a = (m_A + m_B)g*sin(θ) - (μ_Am_A + μ_Bm_B)g*cos(θ)

or

a = g*sin(θ) -  (μ_Am_A + μ_Bm_B)g*cos(θ)/(m_A + m_B)   ...(7)

 

Substitute (7) back into (5) or (6) and rearrange to get T.  I'll leave you to do this and to plug in the numbers (you should get something close to a ≈ 3.5m/s² and T ≈ -25N)