We have \[\frac{1}{(1-x)^{n+1}}=\sum_{i=0}^{\infty}\binom{n+i}{i}x^i.\]Combining \begin{align*} \binom{m+n-2(k+1)}{n} &=\left[x^{m-2(k+1)}\right]\frac{1}{(1-x)^{n+1}} \\ (-1)^k\binom{r}{k} &=\left[x^{2k}\right](1-x^2)^r \end{align*}yields \[p(m, n, r)=\left[x^{m-2}\right]\frac{(1+x)^r}{(1-x)^{n+1-r}}=\sum_{i+j=m-2}\binom{r}{i}\binom{j+n-r}{n-r} > 0.\]For the second part, we have \begin{align*} \sum_{r=0}^{n}p(m, n, r) &= \left[x^{m-2}\right]\frac{1}{(1-x)^{n+1}}\sum_{r=0}^{n}(1-x^2)^r \\ &= \left[x^{m-2}\right]\frac{1}{(1-x)^{n+1}}\frac{1-(1-x^2)^{n+1}}{x^2} \\ &= \left[x^m\right]\left(\frac{1}{(1-x)^{n+1}}-(1+x)^{n+1}\right) \\ &= \binom{m+n}{n}. \end{align*}(There is an extra term $-\binom{n+1}{m}$ which disappears due to $n\leq m-2$.)