On the hills, torque is more important than HP...
14% is 8 degrees.
You need 56 pounds of 'thrust' to maintain a steady speed at an 8 degree grade pushing 400 pounds. On a 26 inch diameter tire, that works out to 61 foot-pounds (60.67...)
Assuming zero losses in your gearing , with a 20:1 gear ratio, the engine must therefore be able to provide 3 foot-pounds of torque.
So, whatever engine you get, with fixed gearing, divide 60.67 time the total drive ratio to calculate the absolute minimum engine torque.
But, if you design a CVT based setup, you could make this work. If the total ratio was 18 to 1, a 2.5:1 CVT reduction at hi load conditions would make the effective gear ratio about 45 to 1.
and, going up hill, dividing 60.67 by 45 yields 1.3 foot pounds. (there are more losses with a CVT though, so you should count on about 1.7 foot pounds engine torque, at a minimum.)
For reference:
The Honda GXH50 is 2.5HP, 2.2 ft-lb max
The Honda GX35 is 1.3 HP, 1.4 ft-lb max
Mitsubishi TLE43 is 2.2 HP, 1.6 ft-lb max