Determine steady state response
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)\$:

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

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.