Voltage across an resistor
It's funny how diode questions are ironically hard, they are often under-looked as simple devices, however they are too much non-linear and impose many paradoxal circuit problems. 

This is rather a very pragmatic way of solving it. When solving diode circuits you use a un-analytical solving method. Assume each combination of conduction and not conduction. 

1) Assume it is conducting. 







simulate this circuit &ndash; Schematic created using CircuitLab

Its pretty straightforward that \$v_o = v_d\$.

However you will find a hidden inequation! 

\$I_3 = I_1-I_2&gt;0\$ this will imply that: 

\$\frac{v_i-v_d}{R_1} - \frac{v_d}{R_2} &gt;0\$ or \$ v_i&gt; v_d (1+\frac{R_1}{R_2})\$

So keep it from now, \$v_0 = v_d\$ if \$ v_i&gt; v_d (1+\frac{R_1}{R_2})\$. 

2) Assume its not conducting, a simple voltage divider and:

\$ v_o = \frac{R_2}{R_1+R_2} v_i\$, however there is another inequation hidden and not \$I_3 &lt; 0\$ this wouldn't help much...

It's \$v_o -v_d &lt; 0\$ or \$v_i &lt;(1+\frac{R_1}{R_2})v_d\$.     YEAH, CONSISTENCY!

We have: \$ v_0 = \frac{R_2}{R_1+R_2} v_i\$ if \$v_i &lt;(1+\frac{R_1}{R_2})v_d\$. 

3) Conclusion: 

\$  v_o = \begin{cases}
v_d,&amp;\quad v_i &gt; (1+\frac{R_1}{R_2})v_d\\
\frac{R_2}{R_1+R_2} v_i,&amp;\quad v_i &lt; (1+\frac{R_1}{R_2})v_d 
\end{cases}\$

Numerically: 

\$  v_o = \begin{cases}
0,6V,&amp; \quad v_i &gt; 1,2V \\
\frac{1}{2} v_i,&amp; \quad v_i &lt; 1,2V
\end{cases}\$

4) Now answering questions: 

a) The output is zero! Yes, because \$v_i &lt; 1 &lt; 1,2\$ always, therefore Vo will be half of Vi always having also 0V as mean.

b) Check previous inequations. 

Edit: After @Granger Obliviate pointed out I did correct the last statement. Thanks!