@lijiwei You must correct (or reformulate) the problem, in particular the definition of Y.
Is h(x) always supposed to nonnegative? If h(x)>0 then s(x)>1, thus (1-s(x)) < 0 so that (1-s(x))^(-1+1/n) = 1/(1-s(x))^(2/7) is imaginary.

@wo0olf Well, then you got the problem that the boundary conditions can only contain derivatives normal to the boundary, thus D[1](T)(x,0)=1000 and D[1](T)(x,0.000025)=2000 are not acceptable.

Could you provide text instead of an image, please.

You have
Y := x->f-(1/2-(1/2)/n)*ln(s(x))+2*ln(1-(1-s(x))^(-1+1/n));
and with the values of your parameters then that evaluates to

thus you have a problem even though you only need diff(Y(x),x):
At the one end x=0 you want h(0)=0. If h(x)>0 for some x then you get an imaginary result because of the last term in Y, try
eval(Y(x),{diff(h(x),x)=1,h(x)=-.1});
If on the other hand h(x)<0 for some x then the fact that you have h(x)^n gives an imaginary result, try
eval(h(x)^n,h(x)=-.1);

The error I get from dsolve is not the one you report, but:
Error, (in fproc) unable to store 'HFloat(HFloat(undefined))+HFloat(HFloat(undefined))*I' when datatype=float[8]

@wo0olf BC1, BC2, and IC already determine a solution:

restart;
pde:= diff(T(x, y), x)-1.775*10^(-5)*(diff(T(x, y), y, y))/(-1.218493440*10^11*y^2+4.244913600*10^6*y+0.33e-1) = 0;
sol:=pdsolve(pde,{T(0, y) = 0,D[2](T)(x, 0) = 1000, D[2](T)(x,0.000025) = 2000},numeric);
sol:-animate(x=1e-3,frames=100);
sol:-plot3d(x=0..1e-3);

@wo0olf This is a new problem: The pde is different and the conditions are different.
You cannot impose conditions like (D[1](T))(0, y) = 1000, (D[1](T))(x, 0) = 2000. The first one because the pde is first order in x, the second because the boundary conditions (as stated in the error message) can only contain derivatives normal to the boundary, thus (D[2](T))(x, 0) = 2000 would have been acceptable.
An example of a problem that can be handled:
sol:=pdsolve(pde,{T(0, y) = 0,D[2](T)(x, 0) = 1000, D[2](T)(x,1) = 2000},numeric);
Maybe I should add that D[1](T) means the derivative of T w.r.t. the first variable (called x in your case). Similarly D[2](T) means the derivative of T w.r.t. the second variable, i.e. y.

@das_goon OK, try this instead:

sys1 := [diff(c(x, t), t) = gDiffusion*10^5*diff(c(x, t), x$2), diff(zeta(x, t), t) = KDiffusion*10^6*diff(zeta(x, t), x$2)];
pds1:=pdsolve(eval(sys1[2],KDiffusion=1),{zeta(0, t) = .4, zeta(x, 0) = .4, D[1](zeta)(0, t) = 0},numeric,time=t,range=0..2000,spacestep=3);
res1:=pds1:-value(output=listprocedure);
Z:=subs(res1,zeta(x,t));
Z(2000,1);
Z(2000,t); #Returns unevaluated as it should
pds2:=pdsolve(eval(sys1[1],gDiffusion=1),{c(2000, t) = Z(2000,t), c(x, 0) = 0, D[1](c)(3000, t) = sin((1/100)*t)},numeric,time=t,spacestep=3);
pds1:-plot3d(t=0..1000); p1:=%:
pds2:-plot3d(t=0..1000); p2:=%:
plots:-display(p1,p2);

It seems that you are just asking if Maple have ways of finding the maximum of a function of a single variable symbolically. Maple can do the usual calculus needed to find a maximum of f(x). Find f'(x), solve f'(x)=0. Examine f''(x). etc. That there are parameters in the function is basically irrelevant. 

@samiyare Yes, I must have made a mistake.
Try this for a start from NBT=2 in the loop in the previous worksheet:
GUESS:=[T(eta) = -eta+.7*eta^2, u(eta) = 0.1e-2, phi(eta) = 0.09, phi0 = 0.09, p2 = 0.4, U(eta) = 0.001*eta, UF(eta) = 0.0001*eta];
LOOP:
res:=GUESS:
for NBT2 in [seq(2-.1*i,i=0..19)] do
  res := dsolve(subs(NBT=NBT2,{ueq2,Teq,eq3,eq4,eq5,U(0)=0,UF(0)=0,((-cbf*rhobf+cp*rhop)*UF(1)+ rhobf*cbf*U(1))/10000=p2,UF(1)=Phiavg*U(1),u(0)=lambda*D(u)(0),u(1)=-lambda*D(u)(1),D(T)(0)=-1,phi(0)=phi0,T(0)=0}), numeric,method=bvp[midrich],approxsoln=res);
  RES[NBT2]:=eval(res)
end do:

I couldn't download the file you attached, but used my own with the new parameters.
You may ask how I got GUESS this time.  I had success with my own unsophisticated code with some wild guesses and constructed GUESS based on that. My code is unsophisticated in the sense that it reports results even if they are lousy. That is sometimes helpful in finding better results.

@Axel Vogt Assuming that all parameters are positive we get
res1:=dsolve(Eq1) assuming positive;
series(rhs(res1),r=0,3); #So _C1 must be zero
diff(eval(%,_C1=0),r); #This is required to be zero

@mehdi jafari It is OK that you can opt out. But to have that as default seems strange to me.

Using the default:
restart;
with(CodeTools):
ode:=2*a*x^3*y(x)^3+2*x*y(x)+diff(y(x), x);
result:=Usage(dsolve(ode=0,y(x)));

you get the result from dsolve and a print of various info including cpu and real time. Do you need the time as actual output?

@mehdi jafari It used to be that if you answered a question or if you made a comment you were automatically signed up for email notification, i.e. subscription by default.

@samiyare Looking at the results obtained for NBT=1 (actually I think I looked at NBT=0.7) I saw that basically the graphs of phi, u, and T look like parabolas. So that is how I got those. You don't have to include the derivatives, as Maple can compute these, but I did initially since I was experimenting with some code of my own, which requires that the system be rewritten as a first order system. If the derivatives are included I assume those are used even if they are not really derivatives of the given approximations. For U and UF I used simple linear approximations.

About the case:

Phiavg:=0.1:
EPSILONE:=0.5:
Nr:=30:
lambda:=0.1;
Ha:=5;
NBT=0.2

I tried with those values, and with nothing else changed. I got no problem for NBT=0.2. In fact no problem from NBT=1 down to NBT=0.08. The first problem occurred at NBT=0.07.

I notice that the following device works:
latex(subs(erf=Erf,r));
and for the other example:
latex(subs(arctanh=Artanh,r));

The error comes up also with this simple version:
latex(erf(2*x));

but not with
latex(erf(x));
Same with arctanh.

The problem may be in `latex/print`, which is different from the procedure with the same name in Maple 17, where there is no such problem.
You may try
`latex/erf`(2*x);
p:=`latex/print`(erf(2*x)); #In Maple 17 p is a symbol, in Maple 18 a sequence
whattype~([p]); #Notice the type `*`
Compare with
p:=`latex/print`(sin(2*x)); #In Maple 17 and 18 p is a sequence
whattype~([p]);

I submitted an SCR about latex(erf(2*x)).