period

i am sorry to tell that the solution of problem 2 given by 霹靂雷霆勁爆ray星 is wrong.
This is because
sin( (x+T)²) = sin(x²)
doesn't imply that there exists an integer k such that
(x+T)² = x² +2kπ
there is another possibility
(x+T)²+x² = Mπ, where m is an odd integer.

Here comes an alternative solution.

Again assume
sin( (x+T)²) = sin(x²) -----(1)
for some fixed positive T and for all real x
put x =0
sin(T²) = 0
so T² = kπ, with k being a positive integer.
(1) implies that
sin( (x+2T)²) = sin(x²) for all real x
since (2T)² = 4kπ = 2k(2π),
sin(x²+4Tx) = sin(x²) for all real x
put x =1 and -1,
sin(1+4T) = sin 1 = sin(1-4T)

case 1:
(1+4T) = 2nπ + (1-4T), where n is an integer
4T = nπ
16kπ = n²π²
impossible since this implies that π = 16k/n², which is rational

case 2:
(1+4T) + (1-4T) = uπ, where u is and odd integer.
then π = 2/u, which is rational. Contradiction

so T doesn't exist.