1) Denote the path as P.
The work done by the non-conservative forces is equal to the change in mechanical energy which here is just mgh.(there is no change in KE)
Let the work done in pulling the block be W.
Then W - \int_P f_r dl = mgh
But f_r = \mu mg \cos \theta where \theta is the angle made by the tangent at a point on the path with the horizontal
\therefore \ \int_L f_r dl = \int_L \mu mg \cos \theta dl = \int_L \mu mg (dl \cos \theta)
But dl \cos \theta = dx where dx is measured along the horizontal (that is the projection of the displacement along the path, taken on the horizontal)
Now its obvious that \int_L dl \cos \theta = \int dx = L
Hence work done by friction = \mu mgL
W = \mu mgL+mgH which is independent of the path up the hill.