HINTS (updated Tue 9:45AM)

Question 2b
===========
Consider an input x (longer than 363 characters), and consider the string 
which is the first 363 character of x. Can you show that if M halts after
 less than 363 steps on x, then it also halts after less than 363 steps on y? 
And how does this help?

 Question 3
==========
Notice that in showing that R(n) is not computable, you cannot assume that 
the only way to compute R(n) is by going over all Turing machines M with n 
states, computing r(M), and taking the maximum. More generally, just like 
showing decidability, it's not a good idea to try to show that certain obvious 
attempts to find machines to decide/compute fail. 
It's about showing the uncomputability for all possible functions.
But how about trying to use a machine that computes f as a subroutine that will
"allow" to decide an undecidable language?

Now, imagine that you knew that a machine cannot run more than a certain # of 
steps before it halts OR that it doesn't halt at all. This would have allowed
you to construct pretty strong machines. Perhaps too strong... Your function 
is not about # of steps before halting, but something slighltly different, so 
there is still some work from this hint to the answer.
 
Question 4
==========
Some people have asked if the answer yes/no should be accompanied by a proof/exaplanation. Of course it should!!
Remember, Rice's thm says that a language of the form 
L = {<M> : L(M) has the property P} is undecidable if P is nontrivial, 
namely it's a property that some machines have and some don't.
An example to a trivial properties is L = {<M> : L(M) recognizable} or
L = {<M> : L(M) is a subset of Sigma*} since these properties always hold, and 
so the corresponding languages are the set of all Turing machines.