Share battery between different voltage devices
The cheapest way is a diode in series with the 3.7 V LiPO. That will drop about 0.6 V (the MSP430 will probably not consume much power). A resistor's voltage drop will be less constant. The MSP430 will operate at voltages as low as 1.8 V, and then the camera will have given up already.   

To simulate a button-push you can control a transistor which you mount parallel to the button, from the MSP430.   

The diode will give you a constant voltage drop, so if the battery's voltage sags to 3.5 V the MSP430 will get 2.9 V instead of 3.1 V. It doesn't matter much for its operation, like I said it will keep working at a voltage as low as 1.8 V. (At work I tested how low it would go, and it still worked at 1.3 V, but that's no longer guaranteed.)

If you want to be a Good Boy you would use a voltage regulator, in this case an LDO, for Low Drop-Out. Many voltage regulators are three pin devices: 1 pin  input, 1 output, and a common ground, that's the negative pole of your voltage. Standard regulators need a few volt more input than output, so for 3 V out you may need 5 V in, more than the battery can supply. An LDO will work with less voltage difference. For instance a 2.5 V LDO may need only 2.7 V input. The regulator will keep the output at 2.5 V as long as the input is higher than the 2.7 V. You'll also need a couple of capacitors, one for the input, one for the output.

Oli's MCP1700 is a good choice. It has a very low ground current, together with the MSP430 you should be able to stay below 10 &micro;A, so you'll barely discharge the battery. 44 cent at Digikey (Farnell is 55 cent). 



A 1N4148 diode costs 10 cent.