The two assertions here are (aside from trivial spelling-modernization) exactly the KJV text of the so-called Johannine Comma (a fragment of John's gospel almost universally reckoned to be a later interpolation, which orthodox Christians might regret since it would be the only explicit statement in the Bible of the doctrine of the Trinity). This means that if there is a puzzle here it's the result of coincidence, divine inspiration, or deliberate foolery by the translators of the KJV; I personally think coincidence the most likely explanation. And this in turn means that we shouldn't expect whatever coincidence forms the foundation of the puzzle to be too impressive :-).
So I propose the following: take "there are three ...: X, Y, and Z; and these three are one" to mean that X=Y=Z=1. Likewise, though it's more of a stretch, for "... and these three agree in one". And take X,Y,Z to denote the sums of their letters, the idea being that each letter gets a distinct integer value. (It would be nice for them to be in the range 1..26, but of course requiring the sums to be 1 prevents that.)
There are indeed lots of solutions. If I have translated the question into Mathematica-ese correctly, without the distinct values requirement we would get a simple 10-dimensional space of solutions: let A,B,D,E,F,G,H,I have any integer values at all, and pick two other arbitrary integers that I'll call x and y; then assign L=1+A+2B+C+2H+3x, O=-B-D-H-x, R=1+A+2B-F+2x, S=A+E+F+H-2I-y, T=-A-B-E-H-x, W=F+H, Y=2B+F-G-2H+2I+x+y. Then we just have to pick our values so that these values are distinct. If, e.g., we take A,B,D,E,F,G,H,I,x,y to be 0,1,2,3,4,5,6,7,8,9 then this is the case: we have L=41, O=-17, P=9, R=15, S=-10, T=-18, W=10, Y=14.
If we remove the "...=1" part, which is after all a bit of a stretch for the second verse, then of course there are a lot more solutions; but perhaps now we can get all the values in the range 1..26? (So that we could, if we wanted, fill them out to give an assignment of numbers 1..26 to the whole Latin alphabet.) Yup: take A,B,D,E,F,G,H,I,L,O,P,R,S,T,W,Y to be 23,18,25,17,2,1,3,8,6,10,22,11,16,4,14,7 respectively.
Of course if we really wanted to take this seriously we'd need to do it in Koine Greek, but let's not.
Taking note of the sixth hint,
if in the first sentence we set A=1 ... Z=26 and give up the ghost so that we have just FATHER, WORD and HOLY then these yield totals of 58, 60, 60 respectively. So if instead we take A=2 ... Z=27 then the totals match up (giving 64 in each case, which is appropriate; 64 is both a square and a cube and I'm sure I've seen someone use the way squares form the boundary of a cube as some sort of analogue of the Christian doctrine of the Trinity...). Perhaps this is what @David has in mind?
This doesn't work at all for the second sentence, though:
WATER and BLOOD yield 67 and 48 respectively and are of the same length.
