Here is the answer to Question 2. It may be probably simplified.

Denote $y=3-x$, then we rewrite your identity as $$\binom{y+K-2}K=\frac{(y-1)y(y+1)\dots (y+K-2)}{K!}=c_0\binom{y}0+c_1\binom{y}1+\dots+c_K\binom{y}K,$$
where $$c_p=p!\sum_{n=p}^K(-K)^{n-p}\frac1{n!}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}}
\sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}.$$
On the other hand, by Vandermonde--Chu identity we have $$\binom{y+K-2}K=\sum_{i=2}^{K}\binom{y}i\binom{K-2}{K-i},$$
so your identity is equivalent to the formula
$$
\sum_{n=p}^K(-K)^{n-p}\frac{K!}{n!}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}}
\sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}=\frac{K!}{p!}\binom{K-2}{K-p},
$$
I multiplied both parts by $K!/p!$. Note that
$$
\frac{K!}{n!}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}}
\sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}
$$
is a number of the trees $T$ on $\{0,1,\dots,K\}$ such that degree of 0 equals $n$ and $p$ vertices in different components of $T\setminus\{0\}$ are marked. Indeed, if these components $A_1,\dots,A_n$ are enumerated (this corresponds to the multiple $n!$) and $i$-th component $A_i$ has $k_i$ vertices, then we have $\frac{K!}{k_1!\dots k_n!}$ ways to choose $A_i$, $\sigma_p(k_1,\dotsc,k_{n}) $ ways to mark $p$ vertices in different components, $k_i^{k_i-1}$ ways to make a tree on $A_i$ and choose a vertex in $A_i$ joined with 0.

Note that each (out of $\binom{K}p$ sets) set of $p$ marked vertices makes the same contribution to the sum. So, we may suppose that the marked set is $\{1,2,\dots,p\}$ and we have to prove that the sum of $(-K)^{n-p}$ over admissible trees (where the tree $T$ is admissible if $1,2,\dots,p$ are in different components of $T\setminus \{0\}$) equals $\frac1{\binom{K}p}\frac{K!}{p!}\binom{K-2}{K-p}=(p-1)p\dots (K-2)$.

We start to prove this from the case $p=0$, $p=1$, where the restriction that $1,2,\dots,p$ are in different components of $T\setminus \{0\}$ disappears. Then the sum $z_0^{n-1}z_1^{d_1-1}\dots z_K^{d_K-1}$, $d_i=\deg(i)$, over all trees on $\{0,\dots,K\}$ equals, as is well known and easy to prove, to $(z_0+\dots+z_K)^{K-1}$. Substituting $z_0=-K$, $z_1=\dots=z_K=1$ we get the result.

Now we deal with the more involved case $p\geqslant 2$. Denote $K=p+m$ and consider the variables $z_0,z_1,\dots,z_p,z_{p+1},\dots$ (infinitely many for simplicity of notations). Denote $s=z_0+z_1+\dots$, write $\sigma_i$ for the $i$-th elementary symmetric polynomial of $z_{p+1},z_{p+2},\dots$. Denote $\varphi_0=1$, $\varphi_m=s\varphi_{m-1}+(p-1)p\dots (p+m-2)\sigma_m$ for $m\geqslant 1$. I claim that the sum of $z_0^{n-p}z_1^{d_1-1}\dots z_{p+m}^{d_{p+m}-1}$ over all admissible trees equals $\varphi_m(z_0,z_1,\dots,z_{p+m},0,0,\dots)$.

Note that this implies our claim, as follows from the substitution $z_0=-K=-p-m,z_1=\dots=z_{p+m}=1$.

The proof is on induction in $m$. Base $m=0$ is clear. For the induction step, look at coefficients of any specific monomial $z_0^{n-p}z_1^{d_1-1}\dots z_{p+m}^{d_{p+m}-1}$. Consider two cases:

1) $d_i=1$ for a certain index $i\in \{p+1,\dots,p+m\}$, without loss of generality $i=p+m$. This corresponds to the case when $p+m$ has degree 1, such a vertex may be joined with any of other vertices, and removing corresponding edge we get a tree (it remains admissible) on $\{0,1,\dots,K-1\}$. This corresponds to the summand $s\varphi_{m-1}$: namely, $z_j\varphi_{m-1}$ corresponds to the edge between $p+m$ and $j$; $j=0,1,\dots,p+m-1$.

2) $d_{p+1},\dots,d_{p+m}$ are greater than 1. Then they are all equal to 2, since the degree of the whole monomial equals $m$. In this case there are $p(p+1)\dots (p+m-1)$ admissible trees (well, they are all admissible for such a choice of degrees and we may either apply the above formula for all trees, or prove it by induction, or as you wish). It remains to prove that the coefficient of $z_{p+1}\dots z_{p+m}$ in the function $\varphi_m$ equals $p(p+1)\dots (p+m-1)$. Since $\varphi_m=s\varphi_{m-1}+(p-1)p\dots (p+m-2)\sigma_m$, it is equivalent to proving that the coefficient of $z_{p+1}\dots z_{p+m}$ in $s\varphi_{m-1}$ equals $p(p+1)\dots (p+m-1)-(p-1)p\dots (p+m-2)=mp(p+1)\dots(p+m-2)$. We should take some $z_j$, $p+1\leqslant j\leqslant p+m$, from the multiple $s=\sum z_i$, and for each choice of $j$ we have a coefficient of $z_j^{-1}\cdot z_{p+1}\dots z_{p+m}$ in $\varphi_{m-1}$ equal to $p(p+1)\dots(p+m-2)$ - by induction (base $m-1=0$ is clear).