Simplification of $\displaystyle{0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$

A combinatorial proof is in order!

The given expression is the number of ways to choose an even-sized subset of $n$ people, and promote one of them to be the leader. Notice that $n\cdot 2^{n-2}$ is the number of ways to choose the leader first, then choose an odd-sized subset from the remaining $n-1$ people (as the number of odd-sized subsets is the same as the number of even-sized ones). Hence the two sides are equal.

Sure. Notice that $$\sum _{i = 0}^{n}2i\binom{n}{2i}=\sum _{i = 0}^{n}\binom{2i}{1}\binom{n}{2i}=\sum _{i = 1}^{n}\binom{n}{1}\binom{n-1}{2i-1}=n\sum _{i=1}^n\binom{n-1}{2i-1}=n\cdot 2^{n-2}.$$ Notice that the last step is because you are adding half of the binomials, and the odd half equals the even half by the binomial theorem.