What does it mean to say an $n \times n$ matrix is Hermitian?

What does it mean to say an $n \times n$ matrix is unitary?

Show that the eigenvalues of a Hermitian matrix are real and that eigenvectors corresponding to distinct eigenvalues are orthogonal.

Suppose that $A$ is an $n \times n$ Hermitian matrix with $n$ distinct eigenvalues $\lambda_{1}, \ldots, \lambda_{n}$ and corresponding normalised eigenvectors $\mathbf{u}_{1}, \ldots, \mathbf{u}_{n}$. Let $U$ denote the matrix whose columns are $\mathbf{u}_{1}, \ldots, \mathbf{u}_{n}$. Show directly that $U$ is unitary and $U D U^{\dagger}=A$, where $D$ is a diagonal matrix you should specify.

If $U$ is unitary and $D$ diagonal, must it be the case that $U D U^{\dagger}$ is Hermitian? Give a proof or counterexample.

Find a unitary matrix $U$ and a diagonal matrix $D$ such that

$$U D U^{\dagger}=\left(\begin{array}{ccc} 2 & 0 & 3 i \\ 0 & 2 & 0 \\ -3 i & 0 & 2 \end{array}\right)$$