First of all: the Fourier transform is a linear operation, which implies
$$\mathcal F\{x+r\}=\mathcal F\{x\}+\mathcal F\{r\}\text.$$
So, when in your sum \$y_1=x+r\$ both \$x\$ and \$r\$ are 0 for any given frequency, then that sum must be zero, too. Thus, the bandlimit for the sum signal is simply the overall swath of spectrum covered by both signals, and since in your case, the support of \$X\$ fits completely in the support of \$R\$, so that the sum signal can only have the bandwidth of \$R\$.
Now, let's look at the time-domain convolution \$y_3(t)=x(t)*r(t)\$; as you correctly noticed,
$$\mathcal F\{x*r\}=\mathcal F\{x\}\cdot\mathcal F\{r\}=X\cdot R\text.$$
We can directly see that if either \$X\$ or \$R\$ are zero for any given point in frequency, than the product must be zero. Thus, since \$X\$ has "more zero", it sets the total bandwidth (make a drawing of \$X(\omega)\$ and \$R(\omega)\$ beneath each other to spot the places where their product can be non-zero).
For the product, things are a bit harder.
I'm not doing the multiplication thing, since your question is definitely homework, and I think I've given you the tools to solve the last, and probably hardest, part.
The main thing I see wrong with your approach is that as drawn, your \$x(t)\$ is actually \$x(t)= u(t)e^{-t}\$, not \$e^{-t}\$. Making this change should remove one the infinite limits on your integration (when the exponential goes to infinity).
Best Answer
Convolution basically tells you how similar two signals are as one of them is shifted in the x-axis and reflected. So consider taking your signal \$g\$ and shifting it by various amounts. The convolution will have a peak when the two signals are mirror images across the y-axis. In this case, they already are, so the peak convolution occurs with a shift of 0.
The most important thing to understand is that the meaning of the x and y axes have changed from (time, value) to (shift, alignment).
I've ignored the meaning of the y-axis magnitude. When I've used the convolution before it's been with normalized signals so that the values ranged from 0 to 1, but that is certainly not always the case.