Solved the first one mathematically, although it turned out to actually be easier than I had thought... For a uniform ring, you only need to consider the x-direction of the force.
So for the contribution of a small unit mass dM, the force dF_x on the mass m can be given by df_x = dF*cos(\theta).
They way I am picturing the ring, \theta is the angle between the direction the force dF is acting and the x-axis.
Using basic trigonometry, cos(\theta) = x / (A^2 + x^2)^1/2.
Also we know dF = GmdM / (A^2 + x^2). therefore you get :
dF_x = (Gmx / ((A^2 + x^2)^3/2)) dM. I f you integrate the LHS you get F_x, and the right hand side is an integral in terms of dM.
Since everything else in the RHS is a constant, this is basically a very easy integral. Note that this solution is much more mathematically rigorous than the previous solution.
Although I generally support strict mathematical practice, as long as you understand the concepts behind the procedure you will be fine.
Hope I was helpfull and good luck with the rest of your homework.