selected solutions to ex2:
Q1 . If L1 is A_TM and L2 is Sigma*, then L1 is recognizable and L2 decidable, 
and yet the intersection (L1) is not decidable. The intersection, however, would
be recognizable as the intersection of two recognizable languages (remember 
that a decidable language is also recognizable).

Q2. Any lnguage is countable. In fact the lexicographic order showis a simple
way to count any language: The first element is the least word in the lexicogrphic orer, the second is the second, and so on.

Q3. Here is a machine R that recognizes HaltSomething:
R = "on input <M>
  (let s1,s2, s3 be the lexicographic order on Sigma*) 
  i=0
  repeat
    i = i+1
      run M on strings s1, s2, s_i for i steps. 
      if one of these runs halted before the ith step ACCEP.
"

This is a simple dovetailing idea. If <M> iis in Haltsomething, then it is 
going to halt on something, say input s_a after b steps. Then in the iteration
 with i = max{a,b}+1 this will be detected and M will be accepted.  
Also, if <M> is not in latsomething, it will never halt prematurely, and so 
R will never accept such an M.
This shws that R is a recognizer for HaltSomething.

Assume we could decide HaltSomething by a decider H. Here is a way to decide Halt_TM by a 
machine D.

D = "on input <M,w>
  run H on M_w (the machine that ignores its input nd runs M on w) and do
the same."

If <M,w> in Halt_TM, then M_w will halt on all inputs, and H will accept <M_w>,
 making D accetpt <M,w>
If <M,w> is not in Halt_TM, then M_w will not halt on any input, and so H will
 reject <M_w> making D reject <M,w>.

But since Halt_TM is not decidable, the existance of D is a contradiction, and
so H does not exst, and HaltSomething is undecidable.