The question asks for a design of a complete system and that's more than I can answer, however I've worked out a rough link budget calculation for your deep space X-band link between Earth orbit and Moon orbit. You can use the same math for each of your UHF links, but if you use a different kind of antenna than a dish you'll have to look up the gains for your UHF antennas elsewhere, the equation below applies only to a circular dish.
For the question of the deep space link we can do that easily by using the math expained in much more detail in this answer:
$$ P_{RX} = P_{TX} + G_{TX} - L_{FS} + G_{RX} $$
- $P_{RX}$: received power
- $P_{TX}$: transmitted power
- $G_{TX}$: Gain of transmitting antenna (compared to isotropic)
- $L_{FS}$: "Free space Loss" but really $\lambda^2 / r^2$ because of the way gain is defined
- $G_{RX}$: Gain of receiving antenna (compared to isotropic)
where
$$L_{FS} = 20 \times \log_{10}\left( 4 \pi \frac{R}{\lambda} \right)$$
and
$$G_{Dish} \sim 20 \times \log_{10} \left( \frac{\pi d}{\lambda} \right).$$
We have antenna diameters of 0.75 meters from the question and let's use 8 GHz as a ballpark/typical deep space X-band frequency. $\lambda = c/f$ gives 0.0375 meters, and that makes the gain of each antenna about 36 dB. The distance to the moon is about 4E+08 meters so $L_{FS}$ is about 223 dB, making $G_{TX} - L_{FS} + G_{RX}$ about -151 dB. That means that for every 1 W of transmit power there will be 8E-15 W of received power.
For an effective receiver temperature of say 300 Kelvin the noise equivalent power or NEP will be about $k_B T \times \Delta f$ where $k_B$ is the Boltzmann constant which is about 1.38E-23 J/K. The required bandwidth will be of the order of the bits per second though the details depend on encoding schemes and error correction outside the scope of this answer.
So with $\Delta f$ of 2E+06 Hz we get an NEP of a 300 K receiver front end of 8E-15 W, which is surprisingly just the same as the received power. That makes the signal to noise ratio $S/N = 1$ and according to the Shannon-Hartley theorem this suggests that yes indeed a bandwidth $BW$ of 2 MHz just barely allow your data rate of 2 Mbit/sec with only 1 Watt of transmit power!
If you use instead 10 Watts and everything else is perfect, you should be okay.
From Am I using Shannon-Hartley Theorem and thermal noise correctly here?:
$$C = BW \ log_2 \left(1 + \frac{S}{N}\right)$$
where $C$ is the theoretical maximum possible data rate.