Let $L$ denote the left-hand side of your identity \eqref{2}. Then, using the identity
$$J_a(x)=\sum_{m\ge0}\frac{(-1)^m}{m!(m+a)!}(x/2)^{2m+a} \tag{$\dagger$}\label{3},$$
we get
$$
\begin{split}
L&=2\sum_{k\ge1}(-1)^k k^3 J_{2k}(x) \\
&=2\sum_{k\ge1}(-1)^k k^3 \sum_{m\ge0}\frac{(-1)^m}{m!(m+2k)!}(x/2)^{2m+2k} \\
&=2\sum_{n\ge1}(x/2)^{2n}\sum_{k=1}^n(-1)^k k^3 \frac{(-1)^{n-k}}{(n-k)! (n+k)!} \\
&=2\sum_{n\ge1}(-1)^n(x/2)^{2n}\sum_{k=1}^n\frac{k^3}{(n-k)! (n+k)!} \\
&=2\sum_{n\ge1}(-1)^n(x/2)^{2n}\frac1{2(n-1)! (n-1)!} \\
&=-\frac{x^2}4\,\sum_{m\ge0}(-1)^m(x/2)^{2m}\frac1{m! m!}
=-\frac{x^2}4\,J_0(x),
\end{split}
$$
by \eqref{3}. This proves your identity \eqref{2}.

Your identity \eqref{1} can be proved similarly. (According to MathOverflow guidelines, there should be only one question in one post.)

To verify the identity
$$\sum_{k=1}^n\frac{k^3}{(n-k)! (n+k)!}=\frac1{2(n-1)! (n-1)!},$$
used in the multi-line display above, one can use the identities
$$
\begin{split}
2k^3 &=((n+k)^{(3)}-(n-k)^{(3)}) \\
&\qquad -(2 - 6 n + 3 n^2)((n+k)^{(1)}-(n-k)^{(1)})
\end{split}
$$
and $(n\pm k)^{(m)}/(n\pm k)!=1/(n-m\pm k)!$, assuming that $\frac{1}{p!}=0$ if $p\in\{-1,-2,\dots\}$,
where $a^{(m)}:=\frac{\Gamma(a+1)}{\Gamma(a-m+1)}=a(a-1)\cdots(a-(m-1))$ for nonnegative integers $m$.

Using the similar identity for $2k^5$ instead of $2k^3$, we can check that
$$
\begin{split}
\sum_{k=1}^n\frac{k^5}{(n-k)! (n+k)!} &=\frac{2n-1}{2(n-1)! (n-1)!} \\
&=\frac{1}{(n-2)! (n-1)!}+\frac{1}{2(n-1)! (n-1)!}
\end{split}.$$