Noisy and Noiseless Channels
In the first channel, you can find $\alpha&gt;0$ by first transmitting a value $x&gt;0$ known by the receiver; then, the receiver can find $\alpha = r/x$. For example, if $x=0.5$, and $r$ is received, then $\alpha = r/0.5$. Once you know $\alpha$, the second symbol allows you to transmit at an infinite rate. Say you want to transmit the bit sequence $b_1\,b_2\,b_3\ldots$. Let $x=0.b_1\,b_2\,b_3\ldots$ in decimal, and you're done. Note that the rate is infinite because you're transmitting an infinite number of bits in two milliseconds: $$\text{Rate} = \frac{\text{information}}{\text{time}} = \infty.$$

For the second channel, note that you can't use that capacity formula, because it assumes that $n_i$ is Gaussian, which is not your case. Rather, you can transmit with zero error probability by defining a set of amplitudes for $x$ such that the noise cannot introduce an error. As a starting point, consider this scenario. You let $x \in \lbrace 0.25, 0.75 \rbrace$, and define a mapping such that a bit zero is transmitted as $x = 0.25$ and a bit one is transmitted as $x = 0.75$. This system has zero errors, since the receiver can apply a decoding rule with threshold $0.5$. It is not possible for the noise to confuse the receiver in this case.

So, what you need to do is figure out how many different values can $x$ take such that the noise cannot introduce errors. That will lead you to the capacity in bits per symbol. Since you know the symbol rate, you can also find the capacity in bits per second.