Let $A$ be a Hermitian matrix ($A^{*}=A$) then, by the spectral theorem there exists an orthonormal basis $\left(e_{i}\right)_{1\leq i\leq n}$ of $\mathbb{C}^{n}$ consisting of eigenvectors of $A$. Each eigenvalue is real. So we have for each $1\leq i\leq n$: $Ae_{i}=\lambda_{i}e_{i}$.
Does that mean a Hermitian matrix has always $n$ distinct eigenvalues? or is it possible that we may get repeated eigenvalues $\lambda_{i}$?
It is possible, that we get repeated eigenvalues. The perhaps simplest example is a real multiple of $\mathrm{id}$. If $\lambda \in \mathbf R$, then $\lambda \mathrm{id}\colon \mathbf C^n \to \mathbf C^n$ is hermitian and has only one ($n$-times repeated) eigenvalue, namely $\lambda$ (and each basis of $\mathbf C^n$ is an eigenbasis).
Note that a linear map $\mathbf C^n\to \mathbf C^n$ with $n$ distinct eigenvalues permits a basis of $\mathbf C^n$ consisting of eigenvectors of $\mathbf C^n$, but only this implication holds. If a map permits an eigenbasis, it need not have $n$ distinct eigenvalues.