Your differential equation is of the second order, but you give three initial data. Also, at the the starting point x==0 of the equation is singular

The error message asked for "initial values for all the dependent variables ", then complained that I gave three I.C.s, even though they are consistent. You are right that x=0 is a singular point, but why should this matter at all? Look at the graphs:

Plot[{ LegendreQ[-(1/2), -1 + 2 x], EllipticK[x]}, {x, -1, 1}, 
 PlotStyle -> {Directive[Lighter@Lighter@Blue, Thick], 
   Directive[Thick, Black, Dashing[.05]]}]

This particular solution is smooth at x=0, so iterative solution algorithms should work fine. In fact, the function EllipticK has a known series expansion, so if nothing else, the NDSolve could expand the series and evaluate. Also compare with:

ICs = MapThread[Equal,
   {(D[T[x], {x, #}] /. x -> 10^(-10)) & /@  Range[0,1], 
    (D[(2/Pi) EllipticK[x], {x, #}] /. x -> 10^(-10)) & /@ 
     Range[0, 1]}];
F = T /. NDSolve[Prepend[ICs, Eq1], 
    T, {x, 10^(-10), 1 - 10^(-15)}][[1]];
LogPlot[F[x] - 2/Pi EllipticK[x], {x, 0, .99}]

Using approximately x=0 as the initial condition, we can compute an accurate but imprecise solution. So it looks like we are getting closer, but we still don't know:

• Is it possible to set the initial condition at x=0 and calculate a numerical solution to arbitrary precision? If yes, how? If no, why not?

Unfortunately, this is starting to look more like the "Mission Critical" failure I was worried about earlier.