Input Current to Output Current for a step-up voltage regulator?
The power formula \$ P = VI \$ gives the relationship between power, voltage and current. 
Your required output is \$ P_{OUT} = 5 \times 1.5 = 7.5 \; W \$.
Assuming you could get a booster with 85% efficiency your input power requirement will be \$ P_{IN} = \frac {7.5}{0.85} = 8.8 \; W \$.
Going back to the power formula your input current at 3.7 V will be \$ I_{IN} = \frac {P_{IN}}{V_{IN}} = \frac {8.8}{3.7} = 2.4 \; A\$.
Run time will be given by \$ hours = \frac {mAh}{mA} = \frac {300}{2400} = \frac {1}{6} = 10 \; minutes\$.


At this point you need to refer to the battery data sheet. The 300 mAh is probably a 10 hour discharge rate - i.e., 30 mA for 10 hours. At higher currents and particularly when exceeding the one-hour discharge rate as you intend to the capacity will be greatly reduced - possibly to about 50%.