Hi there!
(a)
Assuming this is an Atwood machine, we can begin by summing torques about the axis.
Στ = RT2 - RT1
To eliminate the Rs and solve for acceleration, we can rewrite the expression and express alpha as a/r:
Iα = RT2 - RT1
Moment of Inertia of disk = I = 1/2MR²
1/2MR²(a/R) = R(T2 - T1)
Cancel out the Rs:
1/2Ma = T2 - T1
Now, we must sum FORCES:
∑F1 = T1 - m1g
∑F2 = m2g - T2
Rearrange to solve for tension:
T1 = m1a + m1g
T2 = m2g - m2a
Substitute into the above equation:
1/2Ma = m2g - m2a - m1a - m1g
Solve for a:
1/2Ma + m2a + m1a = m2g - m1g

Plug in given values:

a = 1.568 m/s²
(b)
Plug this acceleration into the equation for T1:
T1 = m1a + m1g = .4(1.568) + .4(9.8) = 4.547 N
(c)
Plug the acceleration into the equation for T2:
T2 = m2g - m2a = .6(9.8) - .6(1.568) = 4.939 N