Ideal high pass filter for discrete signal
HINT: Your original LPF is $-\omega_c \lt \omega \lt +\omega_c$. When you shift it by $\pi$, your integral limit for second method is $\int_{-\pi}^{-\pi + \omega_c}$ and $\int_{\pi - \omega_c}^{\pi}$.

For the first method, it should be
$$
h[n] = (-1)^{(n-\alpha)}\frac{sin(\omega_c(n-\alpha))}{\pi(n-\alpha)}
$$
because you have to apply shift of $\alpha$ on $n$. This will fix your plot of (1). 

For (2), with the changed limits as I hinted at the beginning
$$
h[n] = \frac{1}{2\pi}\int_{-\pi}^{-\pi + \omega_c}e^{-j\alpha\omega}e^{j\omega n}d\omega + \frac{1}{2\pi}\int_{\pi-\omega_c}^{\pi}e^{-j\alpha\omega}e^{j\omega n}d\omega\\
= \frac{1}{2\pi j(n-\alpha)}[e^{-j(n-\alpha)(-\pi+\omega_c)} - e^{j(n-\alpha)(-\pi)}] +\frac{1}{2\pi j(n-\alpha)}[e^{-j(n-\alpha)(\pi)} - e^{-j(n-\alpha)(\pi-\omega_c)}] \\
= \frac{1}{2\pi j(n-\alpha)}e^{-j(n-\alpha)(-\pi)}[e^{-j(n-\alpha)\omega_c} - 1]+\frac{1}{2\pi j(n-\alpha)}e^{-j(n-\alpha)(\pi)}[1 - e^{j(n-\alpha)\omega_c}]\\
= \frac{1}{2\pi j(n-\alpha)}(-1)^{(n-\alpha)}[e^{-j(n-\alpha)\omega_c} - e^{j(n-\alpha)\omega_c}]\\
= (-1)^{(n-\alpha)}\frac{sin(\omega_c(n-\alpha)}{\pi(n-\alpha)}
$$