How to prove those "curious identities"?

For the first: $$ \lim_{z=1}\frac{z^n-1}{z-1}=n\tag{1a} $$ $$ \frac{z^n-1}{z-1}=\prod_{k=1}^{n-1}(z-e^{2\pi ik/n})\tag{1b} $$ $$ |1-e^{i2k\pi/n}|=|2\sin(k\pi/n)|\tag{1c} $$ Combining $(1a)$, $(1b)$, and $(1c)$, we get $$ 2^{n-1}\prod_{k=1}^{n-1}\sin(k\pi/n)=n $$ since everything is positive.

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For the second:

If $n$ is even, then $\cos(\frac{\pi}{2})=0$ appears in the product (when $k=n/2$) and $\sin(\frac{n\pi}{2})=0$.

If $n$ is odd, then combining $$ \lim_{z=1}\frac{z^n+1}{z+1}=1\tag{2a} $$ $$ \frac{z^n+1}{z+1}=\prod_{k=1}^{n-1}(z+e^{2\pi ik/n})\tag{2b} $$ $$ 1+e^{i2k\pi/n}=2\cos(k\pi/n)e^{ik\pi/n}\tag{2c} $$ and noting that $\displaystyle\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2}$ so that $\displaystyle\prod_{k=1}^{n-1}e^{ik\pi/n}=(-1)^{(n-1)/2}$ which matches the sign of $\sin(\pi n/2)$, yields $$ 2^{n-1}\prod_{k=1}^{n-1}\cos(k\pi/n)=(-1)^{(n-1)/2}=\sin(\pi n/2) $$

Denote $w = e^{i \pi/n}$. We have

$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \prod_{k = 1}^{n-1} \frac{w^k - w^{-k}}{2i} = \frac{1}{2^{n-1}} \prod_{k = 1}^{n-1} \frac{w^k}{i} (1-w^{-2k})$$

Since we have

$$\sum_{k = 0}^{n-1} x^k = \prod_{k = 1}^{n-1} (x-w^{2k})$$

Setting $x=1$ yields

$$\prod_{k = 1}^{n-1} (1-w^{2k}) = n$$

So we get

$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \frac{n}{2^{n-1}} \frac{w^{n(n-1)/2}}{i^{n-1}} = \frac{i^{n-1}}{i^{n-1}} \frac{n}{2^{n-1}} = \frac{n}{2^{n-1}}$$

I guess (but did not check) that the same kind of reasoning gives the one with $\cos$.

The second purported identity is equivalent to asking for the constant term of $\dfrac{U_{n-1}(x)}{2^{n-1}}$ (i.e., $\dfrac{U_{n-1}(0)}{2^{n-1}}$), where $U_n(x)$ is the Chebyshev polynomial of the second kind. Since

$$\frac{U_{n-1}(x)}{2^{n-1}}=\frac{\sin(n \arccos\,x)}{2^{n-1}\sqrt{1-x^2}}$$

letting $x=0$ gives your identity.