Electronic – Is the frequency response of a circuit only valid for steady state (phasors)

If you replace \$s\$ by \$j\omega\$ you get the system's frequency response \$H(j\omega)\$, which you'll need later on. First you have to compute the Fourier series of the periodic input signal:

$$v_b(t)=\sum_{n=-\infty}^{\infty}c_ne^{jn\omega_0t},\quad \omega_0=\frac{2\pi}{T}$$

where \$T=6\$ms is the period of \$v_b(t)\$. The Fourier coefficients \$c_n\$ are given by

$$c_n=\frac{1}{T}\int_0^Tv_b(t)e^{-jn\omega_0t}dt\tag{1}$$

I haven't evaluated the integral, but it should be pretty straightforward because of all the straight lines in \$v_b(t)\$. Once you have the coefficients \$c_n\$ you need to realize that the response of the system to an exponential input \$e^{j\omega_0t}\$ is simply \$H(j\omega_0)e^{j\omega_0t}\$ (because the system is linear and time-invariant). So you finally get for the output signal

$$v_o(t)=\sum_{n=-\infty}^{\infty}c_nH(jn\omega_0)e^{jn\omega_0t}$$

EDIT: In order to compute the Fourier coefficients you need to write down the piecewise definition of the input signal \$v_b(t)\$:

$$v_b(t)=\left\{\begin{array}{rc}3t-4,& 1\le t<2\\ -3t+8,&2\le t<3\\ -1,&3\le t<7\end{array}\right.$$

Then you split the integral (1) into three intervals:

$$c_n=\frac{1}{6}\left\{\int_1^2(3t-4)e^{-jn\omega_0t}dt+ \int_2^3(-3t+8)e^{-jn\omega_0t}dt- \int_3^7e^{-jn\omega_0t}dt \right\}$$

I guess you can take it from here.

Well, it is just converting the numerator and denominator from complex numbers to magnitude-phase (phasor) notation, which is really short-hand of imaginary exponential.

Numerator

j6 is a vector with no real component (only has imaginary component). So in phasor terms, the magnitude is 6, and the angle is 90° (if it had been negative (-j6), the angle would have been -90°). The angle is measured positive in the anti-clockwise direction, with respect to the positive real axis.

$$ j6 = 6\angle{90°} $$

Denominator

-1-j has real and imaginary components equal to -1. So from the origin it looks like an arrow pointing to the bottom-left in the complex plane. So the angle is -135°, and the magnitude is \$(1^2+1^2)^{1/2} = \sqrt{2} \$

$$ -1-j = \sqrt{2} \angle{-135°} $$

To answer the question in the comments of why \$-j=\frac{1}{j} \$:

When you divide phasors, the resulting magnitude is the quotient of the magnitudes, and the resulting angle is the difference between the angles.

$$ -j = 1\angle{-90} = \frac{1\angle{0}}{1\angle{90}} = \frac{1}{j} $$