Electronic – Linearity and time invariance

I wasn't really able to follow your reasoning, but I can tell you how you can simplify your second equation:

$$9 H(e^{j\pi})$$

Note that $$e^{j\pi} = -1$$ (Euler's formula).

Inserting that into the equation above gives:

$$ \frac{9}{(1-\frac{1}{2}(-1))^2} = \frac{9}{1.5^2}= 4$$ I hope this helps ;)

The system is linear, causal, stable, but not time-invariant.

It is linear because from the definition if \$y_1(t)\$ is the response to input \$x_1(t)\$, and \$y_2(t)\$ is the response to input \$x_2(t)\$, then the response to the input signal \$ax_1(t)+bx_2(t)\$ is given by

$$y(t)=ay_1(t)+by_2(t)$$

It is causal because in order to compute the current output signal, not future values of the input signal are necessary, i.e. to compute \$y(t_0)\$ we only need to know \$x(t_0)\$ (and not even its past values, i.e. the system has no memory).

The system is stable in the BIBO (bounded-input bounded-output sense), because any bounded input signal \$|x(t)|<K\$ produces a bounded output signal \$|y(t)|<L\$ (in our case \$K=L\$).

The system is time-variant because the response to a shifted version of the input signal is not equal to the shifted output signal, i.e. if \$y(t)\$ is the response to \$x(t)\$, then \$y(t-t_0)\$ is generally not the response to \$x(t-t_0)\$. You can see this by noting that the output is always zero for \$-1<t<1\$, no matter how the input signal is shifted. If the system were time-invariant, then also the zero portion of the output would need to be shifted with the input signal.