Transfer function, amplitude response and difference equation for a filter
start from the transfer function: $$
H(z) = \frac{(1-z^{-6})^{2}}{(1-z^{-1})^{2}}
$$

expand it: $$
H(z) = \frac{ 1-2z^{-6} + z^{-12} }{ 1 -2z^-1 + z^{-2}  }
$$

Frequency response is:
$$
H(e^{j\omega}) = \frac{ 1-2e^{-j6\omega} + e^{-j12\omega} }{ 1 -2e^{-j\omega} + e^{-j2\omega}  }$$

Note: this may be simplified by proper algebraic grouping and cancellations to what you have provided.

Transform this rational H(z) directly back into its corresponding LCCDE:
$$y[n]-2y[n-1]+y[n-2] = x[n]-2x[n-6]+x[n-12]$$

I can't figure out a meaning for the wT representation, nor if it is necessary...