If a linear operator $T$ on an inner product space $V$ has eigenvectors which form the basis of $V$, then $T$ is diagonalizable.
If in addition, those eigenvectors are orthogonal, then we have an orthonormal basis of $V$ consisting of eigenvectors of $T$.
What is the advantage of (2) over (1)? Or maybe what are the shared advantages of both (1) and (2) in general?
Also, (2) is possible if and only if $T$ satisfies $T = T^*$ for real inner product space OR $T T^* = T^* T$ for complex inner product space.
Did I get the conditions for (2) right?
If you have an orthogonal basis $\{v_1,\ldots,v_n\}$ and if, for each $k\in\{1,\ldots,n\}$, you define $w_k=\frac{v_k}{\lVert v_k\rVert}$, then $\{w_1,\ldots,w_n\}$. And then if you have an arbitrary vector $u$ of your space, it is very easy to find the coefficients of $u$ in that basis:$$u=\langle u,w_1\rangle w_1+\cdots+\langle u,w_n\rangle w_n.$$
Concerning those final questions, the answer is affirmative in both cases.