For 1: $ \|Tv\|^2=\langle Tv,Tv\rangle =\langle T^*Tv,v\rangle=\langle TT^*v,v\rangle=\|T^*v\|^2. $
For 2, what you say would work if $T$ is invertible, but no one is saying it is. And you wouldn't be using the finite-dimension hypothesis.
If you look at the Schur decomposition of $T$, you have $T=VXV^*$, with $V$ a unitary and $X$ upper triangular. The equality $TT^*=T^2$ implies $XX^*=X^2$.
The diagonal entries of $XX^*$ are non-negative, and they agree with the diagonal entries of $X^2$, which are $X_{kk}^2$ (since $X$ is triangular). So the numbers $X_{kk}^2$ are non-negative, which implies that $X_{kk}$ is real for all $k$. The diagonal entries of $X^2$ are $ X_{11}^2,X_{22}^2,\ldots,X_{nn}^2; $ and the diagonal entries of $XX^*$ are $ X_{11}^2,|X_{12}|^2+X_{22}^2, |X_{13}|^2+|X_{23}|^2+X_{33}^2,\ldots,|X_{11}|^2+\cdots+|X_{1,n-1}|^2+X_{nn}^2. $ So the equality $XX^*=X^2$ implies that $X_{kj}=0$ if $j>k$. That is, $X$ is diagonal with real diagonal, so it is selfadjoint. Then $T$ is selfadjoint.