The solution you have given for the first part is not correct.

Who gives an angle in terms of arctan in Physics? Let's get that out of the way now and say that the angle of incline is theta = 36.87 degrees or so. (I would use a much longer decimal expression for calculations.)

We need to do a Free Body Diagram for each mass. So.

Mass 1: There is a tension T acting on m1 directed up the plane, a normal force N acting perpendicular to the plane, a weight w1 acting directed straight downward, and a friction force f acting down the plane. (We can tell m1 will be moving up the plane since m2 > m1.) My coordinate system will be +x up the plane and +y in the direction of the normal force. (I usually pick one of the directions to be in the direction of the acceleration if there is one. I'll explain why in a minute.)

Now do Newton's 2nd Law in each coordinate direction:

SumFx = T - f - w1*sin(theta) = m1*a

SumFy = N - w1*cos(theta) = 0 (since m1 does not accelerate in this direction.)

Of course we have that w1 = m1*g and f = (mu)*N so these equations become:

T - (mu)N - m1*g*sin(theta) = m1*a

N - m1*g*cos(theta) = 0

We are assuming we know m1 and m2, we know (mu), g, and theta. So this is 2 equations in 3 unknowns. We need another equation to solve the system. So let's move on to m2.

Mass 2: We have a tension T acting upward (equal to the T in the FBD for m1 because we are using an ideal string) and a weight w2 acting downward. I am going to choose a direction for +y to bedownward. Why? Because we chose a positive direction to be up the plane in the m1 FBD and we need to keep the "sign" of the direction of acceleration consistent. If m1 accelerates up and is positive, then m2 accelerates down and should also be positive. (We can choose positive upward here, but that will induce a negative sign on the acceleration in Newton's 2nd Law. My choice avoids this problem.)

Now do Newton's 2nd Law:

SumFy = w2 - T = m2*a (Note the signs on w2 and T! Positive is downward here!)

Again w2 = m2*g so we get

m2*g - T = m2*a

So we have a system of three equations in three unknowns (a, T, and N):

T - (mu)N - m1*g*sin(theta) = m1*a

N - m1*g*cos(theta) = 0

m2*g - T = m2*a

We can immediately solve the middle equation for N and substitute that in the top equation:

N = m1*g*cos(theta)

T - (mu)*m1*g*cos(theta) - m1*g*sin(theta) = m1*a

m2*g - T = m2*a

We want an expression for a, so I am going to solve the bottom equation for T and substitute that into the top equation:

T = m2*g - m2*a

m2*g - m2*a - (mu)*m1*g*cos(theta) - m1*g*sin(theta) = m1*a

m2*g - (mu)*m1*g*cos(theta) - m1*g*sin(theta) = m1*a + m2*a

We can factor the common "a" on the RHS and then divide both sides of the equation by m1 + m2:

a = [m2*g - m1*g*{ (mu)*cos(theta) + sin(theta) }]/(m1 + m2)

-Dan