Question about ramp filter used in filtered backprojection
To see if your math is correct, it is useful to first understand what is, in general, the effect of a filter on a signal, and then see if you can predict what the theoretical result should look like.

If a filter has frequency response $H(Q)$, this means that its response to an input $e^{j2\pi Q_0 r}$ is the signal $H(Q_0)e^{j2\pi Q_0 r}$. In other words, a sinusoidal input of frequency $Q_0$ produces an output of the same frequency, but with amplitude $|H(Q_0)|$ and phase $\angle H(Q_0)$.

In your first question, the input has frequency $Q_0=0$. The filter's response at that frequency is $|Q_0|=0$. Then, the filter's output will be 0: frequency $Q_0=0$ is completely absorbed by the filter and it does not appear at the output.

In your second question, the input has frequency $Q_0=w/2\pi$. The filter's response at that frequency is $|Q_0|=w/2\pi$. The output, then, should be $\frac{w}{2\pi}\sin(wr)$.

As you can see, following this line of reasoning we obtain the same results as you did by just solving the equations. This corroboration should give you confidence that your results are correct, and most importantly, you can understand what the filter is doing physically.