@mmcdara 

You missed 13 cycles.

{[1,3,5],[1,3,6],[1,4,5],[1,4,6],[2,3,5],[2,3,6],[2,4,5],[2,4,6],[1,3,2,4],[1,3,2,5],[1,3,2,6],[1,3,5,4],[1,3,6,4],[1,4,2,5],[1,4,2,6],[1,5,2,6],[1,5,3,6],[1,5,4,6],[2,3,5,4],[2,3,6,4],[2,5,3,6],[2,5,4,6],[3,5,4,6],[1,3,2,4,5],[1,3,2,4,6],[1,3,2,5,4],[1,3,2,6,4],[1,3,5,2,4],[1,3,5,2,6],[1,3,5,4,6],[1,3,6,2,4],[1,3,6,2,5],[1,3,6,4,5],[1,4,2,3,5],[1,4,2,3,6],[1,4,5,2,6],[1,4,5,3,6],[1,4,6,2,5],[1,4,6,3,5],[1,5,2,3,6],[1,5,2,4,6],[1,5,3,2,6],[1,5,4,2,6],[2,3,5,4,6],[2,3,6,4,5],[2,4,5,3,6],[2,4,6,3,5],[1,3,2,5,4,6],[1,3,2,6,4,5],[1,3,5,2,4,6],[1,3,5,2,6,4],[1,3,5,4,2,6],[1,3,6,2,4,5],[1,3,6,2,5,4],[1,3,6,4,2,5],[1,4,2,5,3,6],[1,4,2,6,3,5],[1,4,5,2,3,6],[1,4,5,3,2,6],[1,4,6,2,3,5],[1,4,6,3,2,5],[1,5,3,2,4,6],[1,5,4,2,3,6]}

@weidade37211 

When working with such "extreme" numbers a good idea is to transform the equation symbolically.
 

@Carl Love 

Here is the winding number version. It should work for any closed curve.

ContourInt:= proc(
   f::algebraic, 
   C::(name= algebraic), 
   trange::name= range(realcons)
)
local z:= lhs(C), t:= lhs(trange), r:= rhs(C), eq, R;
R:=remove(z-> type(rhs(z), infinity), op~({singular(f,z)} ));
2*Pi*I*add(round(evalf(Int(diff(r,t)/(r-eval(z,eq)),trange)/(2*Pi*I)))*residue(f,eq),eq=R);
end proc:

@Carl Love 

To decide whether a singular point is in the interior we can compute its winding number (it's enough to approximate and apply round).

@Lali_miani 

You already got the explanation: sqrt(2*x+1)/(2*x+1) is automatically simplified to 1/sqrt(2*x+1). The automatic simplifications cannot be prevented, e.g. you cannot see 2/10 instead of 1/5. There are some tricks, but I think that it is much more important to focus on maths!

@Carl Love 

This is probably the most efficient method.
But I think that for a beginner, a more useful construction (to be often used) is

MyMatrix := Matrix( 2,3, (i,j) -> elem[i,j](t) );

@dharr 

So, I'd formulate:

When x is a sparse Matrix, Vector or rtable, for efficiency reasons, only the nonzero entries are scanned and the zero entries are then appended.

@Kitonum 

evalc does not work well if Zeta is present.

@Al86 

It is easy to compute by hand if you know elementary Calculus. E.g. the limit of the numerator is:
exp(-t) - exp(-infinity)  = exp(-t).

@Al86 You can try (e.g. for t=2)

Student[Calculus1][LimitTutor]();

but AFAIK it works for (very) simple limits.

@digerdiga 

In the future it would be nice to mention the method(s) you know about for your problem.
Otherwise, the answers seem to be useless because you knew them.

@Carl Love 

Nice and professional (as always :-)).

@digerdiga 

asympt cannot switch branches. You can obtain the same result without puiseux.

restart;
sol:=RootOf((8*n-8)*_Z^6+(n^4+36*n^2-68*n+56)*_Z^5+(n^5+10*n^4+80*n^3-200*n^2+224*n-152)*_Z^4+(n^6+28*n^5+69*n^4-268*n^3+468*n^2-356*n+200)*_Z^3+(3*n^7+32*n^6+7*n^5-204*n^4+380*n^3-544*n^2+272*n-128)*_Z^2+(3*n^8+14*n^7-20*n^6-32*n^5+252*n^4-240*n^3+304*n^2-80*n+32)*_Z-n^9-12*n^8-44*n^7-40*n^6-4*n^5-128*n^4+48*n^3-64*n^2):
f:=eval(op(sol), [_Z=z,n=1/x]):
g:=numer(eval(f, z=a/x+b+c*x)):
S:=solve({seq(coeff(g,x,k),k=0..2)}):
subs(S[],x=1/n, a/x+b+c*x);

@Kitonum 

I know. SymFun just calls solve/identity which could (and should) be improved a lot (you are free to send a SCR for it, and if successful, SymFun will work for your examples). Actually, if we are interested only in an odd/even result, this is easy to obtain.
BTW, the "hack" part was included just to answer (c). SymFun can be of course improved, but I prefer a capable solve/identity.

@Carl Love 

To find the initial point of a p-cycle simply do:

f:=y-> 4*y*(1-y):
p:=2;
solve((f@@p)(x)=x); # or fsolve