Calculating the number of elements of a given order in Dihedral group.

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I am studying the structure of $D_n$, the Dihedral group of order $2n$, mainly for competitive exams. I found somewhere that:

$1.$ Number of elements of order $2$ in $D_n$ is $\begin{cases}n,\text{ if } n \text{ odd}\\ n+1,\text{ if } n \text{ even} \end{cases}$

$2.$ For any divisor $d$ of $n$, there is an element of order $d$.

$3.$ Number of elements of order $d\mid n$ and $d\neq 2$ is $\varphi(d)$, where $\varphi$ is the Euler totient function.

What I want are the proofs of these results. I am a bit weak in Dihedral groups. So, can someone help me by providing proofs or at least hints for the proof?

Addendum

I understand the claim $(1)$ because $D_n$ is composed of rotations $1,r,r^2,...,r^{n-1}$ and reflections $f,fr,...,fr^{n-1}$ and the reflections are of order $2$, so there are at least $n$ elements of order $2$. Now the only possiblity of order $2$ element among rotations is $r^{n/2}$ which exists iff $n$ is even. So, $(1)$ is proved. I also understand $(2)$ and $(3)$ because for $d|n$ and $d\neq 2$,elements of order $d$ are elements of $\{1,r,r^2,...,r^{n-1}\}$ which is a cyclic group of order $n$,so elements of order $d\mid n$ exists for $d\neq 2$ and they are $\varphi(d)$ in number.

Is this correct?

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Well for $2)$, take the rotation $r$ of order $n$. It generates a cyclic subgroup $\langle r \rangle$ of order $n$. So the result is a property of cyclic groups.

For $3)$, look at that same cyclic group. A well known property is that there are $\varphi(d)$ elements of order $d$ for each divisor $d$ of $n$.