While I haven't found a proof that #sqrt(2)^(sqrt(2)^sqrt(2))# is irrational, here are answers to the first and third parts.
#sqrt(2)^sqrt(2)# is irrational:
The Gelfond-Schneider theorem states that given algebraic numbers #a, b# where #a != 0, 1# and #b# is irrational, #a^b# is transcendental.
As #sqrt(2)# is a root of #x^2-2#, it is algebraic. It is well known that #sqrt(2)# is irrational, although this fact may be proven by supposing that #sqrt(2)=p/q# with #p, q in ZZ#, squaring both sides, multiplying by #q^2#, and then noting that the left hand side will have an odd exponent for #2# in its prime factorization, while the right hand side will have an even exponent, a contradiction.
By the above, #sqrt(2)^sqrt(2)# fulfills the conditions for the theorem, and thus is transcendental (and therefore irrational).
#sqrt(2)^(sqrt(2)^(sqrt(2)^(...))) = ""^oosqrt(2)# is rational:
We will prove the stronger result that #""^oosqrt(2)=2#.
First, to show that #""^oosqrt(2)# converges, we will show that the sequence #a_n = ""^nsqrt(2)# is monotone increasing and bounded above.
We proceed by induction. As our base case, note that
#a_1 = sqrt(2) < sqrt(2)^sqrt(2) = a_2 < sqrt(2)^2 = 2#. Now, suppose that for some positive integer #k#, the sequence #a_1, a_2, ..., a_k# is increasing and #a_k < 2#. Then
#a_k = sqrt(2)^(a_(k-1)) < sqrt(2)^(a_k) = a_(k+1) < sqrt(2)^2 = 2#
Thus, by induction, #a_n# is increasing and bounded above by #2#.
Now that we know #""^oosqrt(2)# converges, let #a= ""^oosqrt(2)#. Then #sqrt(2)^a = a#, meaning #a# is a root of #f(x) = sqrt(2)^x - x#. Finding the critical points of #f(x)#, we have
#f'(x) = 1/2ln(2)sqrt(2)^x - 1#
#=> f'(x) = 0 <=> x = log_sqrt(2)(2/ln(2))#
As #f(x)# has only one critical point, it can have at most two roots. By observation, #2# and #4# are both roots of #f(x)#, meaning #a in {2, 4}#. But, by the above work, #a<=2#, meaning #a=2#.
