For the first, use the ]Cauchy-Schwarz inequality, $\displaystyle ||x+y||\le ||x||+ ||y||$, and its consequence, $\displaystyle ||x- y||\ge ||x||- ||y||$.
The second looks straight forward. First, since T is 1-1 and onto, it has and inverse, a function $\displaystyle T^{-1}$, fro Y to X such that $\displaystyle T^{-1}(T(x))= x$ for all x in X and $\displaystyle T(T^{-1}(y))= y$ for all y in Y. Now, suppose u and v are in X and let p= T(u), q= T(v). What can you say about $\displaystyle T^{-1}(p+ q)= T^{-1}(T(u)+ T(v))$?