How to produce a high-pass filter from a low-pass one?
The result will indeed be a high pass filter. From your difference equation, the transfer function of the low pass filter is

$$H_l(z)=\frac{\beta}{1-(1-\beta)z^{-1}}\tag{1}$$

with $\beta=1/\alpha$.
Note that this is actually a leaky integrator, not a classic low pass filter, because its frequency response does not have a zero at Nyquist.

The high pass filter has the transfer function

$$H_h(z)=1-H_l(z)=(1-\beta)\frac{1-z^{-1}}{1-(1-\beta)z^{-1}}\tag{2}$$

which has a zero at DC (i.e. at $z=1$), as it should. However, it is scaled, i.e. its value at Nyquist ($z=-1$) is not $1$, but $2(1-\beta)/(2-\beta)$. So in order to have a 0 dB gain at Nyquist, you need to scale your high pass filter with the inverse of that factor:

$$H'_h(z)=\frac{2-\beta}{2(1-\beta)}H_h(z)=\left(1-\frac{\beta}{2}\right)\frac{1-z^{-1}}{1-(1-\beta)z^{-1}}\tag{3}$$

EDIT:

You can compute the 3dB cut-off frequencies of the low pass filter (1) and of the high pass filter (2) by solving

$$|H_l(e^{j\omega_c})|^2=\frac12\quad\text{and}\quad |H'_h(e^{j\omega_c})|^2=\frac12$$

After some algebra this gives for the low pass filter

$$\cos\omega_c=1-\frac{\beta^2}{2(1-\beta)}$$

and for the high pass filter

$$\cos\omega_c=\frac{1}{1+\frac{\beta^2}{2(1-\beta)}}$$

where $\omega$ is the normalized frequency in radians:

$$\omega=2\pi\frac{f}{f_s}$$

with the sampling frequency $f_s$.