(1)

オイラー多項式の定義より、
\begin{align*} E_{n}\left(1-x\right) & =\sum_{k=0}^{n}C\left(n,k\right)\frac{E_{k}}{2^{k}}\left(\left(1-x\right)-\frac{1}{2}\right)^{n-k}\\ & =\sum_{k=0}^{n}C\left(n,k\right)\frac{E_{k}}{2^{k}}\left(-x+\frac{1}{2}\right)^{n-k}\\ & =\sum_{k=0}^{n}C\left(n,k\right)\frac{E_{k}}{2^{k}}\left(-1\right)^{n-k}\left(x-\frac{1}{2}\right)^{n-k}\\ & =\left(-1\right)^{n}\sum_{k=0}^{n}\left(-1\right)^{k}C\left(n,k\right)\frac{E_{k}}{2^{k}}\left(x-\frac{1}{2}\right)^{n-k}\\ & =\left(-1\right)^{n}\sum_{k=0}^{n}\left(-1\right)^{2k}C\left(n,2k\right)\frac{E_{2k}}{2^{2k}}\left(x-\frac{1}{2}\right)^{n-2k}\\ & =\left(-1\right)^{n}\sum_{k=0}^{n}C\left(n,2k\right)\frac{E_{2k}}{2^{2k}}\left(x-\frac{1}{2}\right)^{n-2k}\\ & =\left(-1\right)^{n}\sum_{k=0}^{n}C\left(n,k\right)\frac{E_{k}}{2^{k}}\left(x-\frac{1}{2}\right)^{n-k}\\ & =\left(-1\right)^{n}E_{n}\left(x\right) \end{align*} となるので与式は成り立つ。

(2)

\(n\in\mathbb{N}_{0}\)のとき、
\begin{align*} E_{2n}^{\left(2j\right)}\left(0\right) & =\left[\frac{d^{2j}}{dx^{2j}}E_{2n}\left(x\right)\right]_{x=0}\\ & =\left[\frac{d^{2j}}{dx^{2j}}\sum_{k=0}^{2n}C\left(2n,k\right)\frac{E_{k}}{2^{k}}\left(x-\frac{1}{2}\right)^{2n-k}\right]_{x=0}\\ & =\left[\sum_{k=0}^{2n}C\left(2n,k\right)\frac{E_{k}}{2^{k}}P\left(2n-k,2j\right)\left(x-\frac{1}{2}\right)^{2n-k-2j}\right]_{x=0}\\ & =\sum_{k=0}^{2n}C\left(2n,k\right)\frac{E_{k}}{2^{k}}P\left(2n-k,2j\right)\left(-\frac{1}{2}\right)^{2n-k-2j}\\ & =\sum_{k=0}^{2n}\left(-1\right)^{k}C\left(2n,k\right)E_{k}P\left(2n-k,2j\right)\left(-\frac{1}{2}\right)^{2n-2j}\\ & =\left(-\frac{1}{2}\right)^{2n-2j}\left(2j\right)!\sum_{k=0}^{2n}\left(-1\right)^{k}C\left(2n,k\right)C\left(2n-k,2j\right)E_{k}\\ & =\left(-\frac{1}{2}\right)^{2n-2j}\left(2j\right)!\sum_{k=0}^{2n}\left(-1\right)^{k}C\left(2n,2n-2j\right)C\left(2n-2j,k\right)E_{k}\\ & =\left(-\frac{1}{2}\right)^{2n-2j}\left(2j\right)!C\left(2n,2n-2j\right)\sum_{k=0}^{2\left(n-j\right)}\left(-1\right)^{k}C\left(2\left(n-j\right),k\right)E_{k}\\ & =\left(-\frac{1}{2}\right)^{2n-2j}\left(2j\right)!C\left(2n,2n-2j\right)\sum_{k=0}^{2\left(n-j\right)}C\left(2\left(n-j\right),2k\right)E_{2k}\\ & =\left(-\frac{1}{2}\right)^{2n-2j}\left(2j\right)!C\left(2n,2n-2j\right)\delta_{0,n-j}\cmt{\delta_{0,j}=\sum_{k=0}^{j}C\left(2j,2k\right)E_{2k}}\\ & =\left(2n\right)!C\left(2n,2n-2j\right)\delta_{n,j} \end{align*} より、
\begin{align*} E_{2n}\left(x\right) & =\sum_{j=0}^{\infty}E_{2n}^{\left(j\right)}\left(0\right)\frac{x^{j}}{j!}\\ & =\sum_{j=0}^{\infty}E_{2n}^{\left(2j\right)}\left(0\right)\frac{x^{2j}}{\left(2j\right)!}+\sum_{j=0}^{\infty}E_{2n}^{\left(2j+1\right)}\left(0\right)\frac{x^{2j+1}}{\left(2j+1\right)!}\\ & =\sum_{j=0}^{\infty}\left(2n\right)!C\left(2n,2n-2j\right)\delta_{n,j}\frac{x^{2j}}{\left(2j\right)!}+\sum_{j=0}^{\infty}E_{2k}^{\left(2j+1\right)}\left(0\right)\frac{x^{2j+1}}{\left(2j+1\right)!}\\ & =x^{2n}+\sum_{j=0}^{\infty}E_{2k}^{\left(2j+1\right)}\left(0\right)\frac{x^{2j+1}}{\left(2j+1\right)!} \end{align*} となる。
これより、
\begin{align*} E_{2n}\left(x\right)+E_{2n}\left(-x\right) & =x^{2n}+\sum_{j=0}^{\infty}E_{2k}^{\left(2j+1\right)}\left(0\right)\frac{x^{2j+1}}{\left(2j+1\right)!}+\left(-x\right)^{2n}+\sum_{j=0}^{\infty}E_{2k}^{\left(2j+1\right)}\left(0\right)\frac{\left(-x\right)^{2j+1}}{\left(2j+1\right)!}\\ & =x^{2n}+\sum_{j=0}^{\infty}E_{2k}^{\left(2j+1\right)}\left(0\right)\frac{x^{2j+1}}{\left(2j+1\right)!}+x^{2n}-\sum_{j=0}^{\infty}E_{2k}^{\left(2j+1\right)}\left(0\right)\frac{x^{2j+1}}{\left(2j+1\right)!}\\ & =2x^{2n} \end{align*} となる。
この両辺を\(x\)で微分すると、\(n\in\mathbb{N}\)のとき、
\[ 2nE_{2n-1}\left(x\right)-2nE_{2n-1}\left(-x\right)=2\cdot2nx^{2n-1} \] となり、
\[ E_{2n-1}\left(x\right)-E_{2n-1}\left(-x\right)=2x^{2n-1} \] となる。
まとめると、
\[ \begin{cases} E_{2n}\left(x\right)+E_{2n}\left(-x\right)=2x^{2n} & n\in\mathbb{N}_{0}\\ E_{2n-1}\left(x\right)-E_{2n-1}\left(-x\right)=2x^{2n-1} & n\in\mathbb{N} \end{cases} \] となるのでこれを1つにすると、\(n\in\mathbb{N}_{0}\)のとき、
\[ E_{n}\left(x\right)+\left(-1\right)^{n}E_{n}\left(-x\right)=2x^{n} \] となり与式は成り立つ。

(3)

(2)より、
\begin{align*} \sum_{k=1}^{m}\left(-1\right)^{k}k^{n} & =\sum_{k=1}^{m}\left(-1\right)^{k}\frac{1}{2}\left(\left(-1\right)^{n}E_{n}\left(-k\right)+E_{n}\left(k\right)\right)\\ & =\frac{1}{2}\sum_{k=1}^{m}\left(-1\right)^{k}\left(E_{n}\left(1+k\right)+E_{n}\left(k\right)\right)\\ & =-\frac{1}{2}\sum_{k=1}^{m}\left(\left(-1\right)^{1+k}E_{n}\left(1+k\right)-\left(-1\right)^{k}E_{n}\left(k\right)\right)\\ & =-\frac{1}{2}\left(\left(-1\right)^{1+m}E_{n}\left(1+m\right)+E_{n}\left(1\right)\right)\\ & =\frac{1}{2}\left(\left(-1\right)^{m}E_{n}\left(1+m\right)-E_{n}\left(1\right)\right) \end{align*} となるので与式は成り立つ。