Using the system function $H(z)$ to the solve the output of a DT LTI system
Let's first briefly recall the meaning of $H(z)$.
Let $y[n]$ be the output of the LTI system. Given an input $x[n]$, the output can be written
$$y[n] = h[n] *x[n] = \sum_{k=-\infty}^\infty h[k]x[n-k]$$
where $h[n]$ is the impulse response of the LTI system en $*$ represents the convolution. Let's now assume that $x[n]$, the input of the LTI system, is of the form $z^n$. In this case,
$$y[n] = \sum_{k=-\infty}^\infty h[k]z^{n-k} = z^n\left(\sum_{k=-\infty}^\infty h[k]z^{-k}\right) = z^nH(z)$$
where $H(z)$ denotes the transfer function of the LTI system (the z-transform of the impulse response). It can be seen that that $z^n = e^{j\Omega}$ is an eigenfunction of the system, that is, it is only affected by a complex factor (i.e. $H(z)$, the so-called eigenvalue) at the output of your system.



Now, your input $f[n]$ can be written as
$$f[n] = \mathcal{R}\{Ae^{j(\frac{\pi}{2}n + \frac{2\pi}{10})}\} = \mathcal{R}\{Ae^{j\frac{2\pi}{10}}e^{j\frac{\pi}{2}n}\}$$
and $Ae^{j\frac{2\pi}{10}}e^{j\frac{\pi}{2}n}$ matches the form of an eigenfunction (with a complex factor $Ae^{j\frac{2\pi}{10}}$). The output of your system to $Ae^{j\frac{2\pi}{10}}e^{j\frac{\pi}{2}n}$ is then given, using linearity of your system and the polar notation of $H$, by
$$y[n] = Ae^{j\frac{2\pi}{10}}H(e^{j\frac{\pi}{2}})e^{j\frac{\pi}{2}n} = Ae^{j\frac{2\pi}{10}}|H(e^{j\frac{\pi}{2}})|e^{\arg\{H(e^{j\frac{\pi}{2}})\}}e^{j\frac{\pi}{2}n}.$$
Finally, again by linearity, the output of your system to $f[n]$ is given by the real part of that.