Filter H(z) manipulation frequency response changes
It is just like ordinary functions manipulation.
You have to remember that z and the frequency are linked with $z = exp(j \omega \cdot T)$
$H(z^2)$ you are doing something equivalent to $H(2 \omega)$ on the continuous frequency domain, i.e. you are narrowing the filter band, and thus stretching its time response.
On the second example, you have to remember that $H_1(z) H_2(z)$ is the response of two cascading filter, that the product of the Z transforms is the convolution of the the time responses.
The filter you are suggesting $0.5 (1 - z^{-1})$ will be a sort of derivative estimate because a signal with response $x[n]$ will be transformed to $(x[n] - x[n-1])/2$ and this is a high-pass filter. You can also think of it as the inverse of the step response, you can spend some time contemplating a table of pairs of signals and their Z transform, for instance this and slowly but certainly you will build some intuition.