I'll sketch the idea behind your claims, starting with Problem 1.

**Step 1:** Convince yourself that if we denote
$$F_p(j):=\frac1j\sum_{i=0}^{j-1}\left(\frac{j}2-i-1\right)^p$$
then $F_p(j)$ is always an **even** polynomial in $j$; a polynomial in $j^2$. For example, when $p$ is odd, we get $F_p(j)=-\left(\frac{j}2\right)^{p-1}$.

**Step 2:** Therefore, from Step 1, it suffices to consider even powers of $j$ instead of $F_p(j)$. We shall also disregard $\frac{-2}{\binom{2m}m}$ and focus on the sum
$$G_p(m):=\sum_{j=1}^m(-1)^j\binom{2m}{m-j}j^{e}$$
where $e\geq0$ is a fixed even integer. Observe that
$$\sum_{j=-m}^m(-1)^j\binom{2m}{m-j}j^e
=\begin{cases} 2G_p(m) \qquad \qquad \text{if $e\neq0$} \\
2G_p(m)+1 \qquad \,\,\text{if $e=0$}.\end{cases}$$
**Step 3:** After re-indexing $k=m-j$,
$$\sum_{j=-m}^m(-1)^j\binom{2m}{m-j}j^e=(-1)^m\sum_{k=0}^{2m}(-1)^k\binom{2m}k(m-k)^e.$$
It's well-known (actually the crux of the matter here) that if $n>a$ then we've the vanishing of
$$\sum_{k=0}^n(-1)^k\binom{n}k k^a=0.\tag{0}$$
The reason why you get a persistent value (for $m\geq m_p$) is caused by the one single term when $e=0$ which happens for $p$ even.

At any rate, we gather that if $e$ is even and $m\geq \frac{e}2$ then
$$G_p(m)=\begin{cases} \,\,\,\,\,0 \qquad \,\text{when $e\neq0$} \\
-\frac12 \qquad \text{when $e=0$}.
\end{cases}$$

To understand Problem 2, it remains to identify the constant term in $F_p(j)$ w.r.t. the variable $j$. To this end, the Binomial Theorem furnishes
\begin{align} F_p(j)
=\frac1j\sum_{i=1}^j\sum_{k=0}^p\binom{p}k\left(\frac{j}2\right)^{p-k}(-1)^ki^k
=\sum_{k=0}^p\binom{p}k\left(\frac{j}2\right)^{p-1-k}(-1)^k\sum_{i=1}^ji^k
\tag{1}
\end{align}
revealing that the only way to encounter the sought-after constant term is provided that $k=p$. Keeping in mind that $\sum_{i=1}^ji^k$ is divisible by $j$, we just look at the quantity (using know expressions for Bernoulli numbers)
$$\frac{(-1)^p}j\sum_{i=1}^ji^p=(-1)^p\left[j^{p-1}+\frac1{p+1}\sum_{r=0}^p\binom{p+1}rB_rj^{p-r}\right].\tag{2}$$
Thus, the term we seek is $(-1)^pB_p=B_p$ (remember: $p$ is even) leading to
\begin{align}
\frac2{\binom{2m}m}\sum_{j=1}^m(-1)^{j-1}\binom{2m}{m+j}F_p(j)
&=\frac{2B_p}{\binom{2m}m}\sum_{j=1}^m(-1)^{j-1}\binom{2m}{m+j} \\
&=\frac{2B_p}{\binom{2m}m}\frac12\binom{2m}m \\
&=B_p
\end{align}
which is exactly what we want to arrive at. We've already known from above the $B_p=0$ when $p$ is odd. The proof is complete for both problems.