A really easy one. Make each line an assignment to a verb and evaluate it at the end. I'll add a rank ("_) to the last word to make sure they are all verbs, and use string replacement to J syntax.
require 'strings' 0: ([: ". '"_' ,~ ]);._2 ('/';'%';':';'=:') stringreplace LF ,~ wd 'clippaste' 0 root'' 152
The 0 was just to eat up the blank lines that are the result when (". y) produces a verb result.
Part 2 is just as easy. I redefine humn as ] so that the expression it appears in is not constant.
humn =: ] root pppw + sjmn pppw'' sjmn'' 150
pppw contains the ] and evaluates to empty on an empty argument, while sjmn has all scalar constant verbs and evaluates to a constant. I want to know what to put in for the ] to get the same constant value out of pppw. In other words, I want to apply the inverse of pppw.
pppw^:_1 sjmn'' 301
I'm glad I was using J to solve this.