DTFT of even and odd samples

Question

Here to find DTFT of $h(2n)$ they have scaled omega, while in RHS to find DTFT $x(2n+1)$ they didn't, why is that?

Answer 1

Because they made a mistake. The very first equation on the right-hand side is wrong. It should be

$$\textrm{DTFT}\big\{x[2n+1]\big\}=\sum_{n=-\infty}^{\infty}x[2n+1]e^{-jn\omega}=\sum_{n\textrm{ odd}}x[n]e^{-j(n-1)\omega/2}\tag{1}$$

Using the trick they suggested, $(1)$ can be written as

$$\textrm{DTFT}\big\{x[2n+1]\big\}=\frac{e^{j\omega /2}}{2}\left[\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega/2}-\sum_{n=-\infty}^{\infty}(-1)^nx[n]e^{-jn\omega/2}\right]\tag{2}$$