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(a) consider computing P(e) using variable elimination
First we have factors
f(a)
f(b)
f(abc)
f(bd)
f(ec)
f(fc)

We then sum out a which eliminates f(A) and f(ABC)
generating f(BC) where p(c|b) = 0.1(0.9) + 0.9(0.3) = 0.36
and p(c|¬b) = 0.1(0.8) + 0.9(0.4) = 0.44

We then sum out f which eliminates f(fc) generating f(c).
we then sum out d which eliminates f(db) generating f(b).

we then sum out b which eliminates f(bc) and f(b), generating f(c) where
p(c)= 0.8(0.36) +0.2(0.44) = 0.376

We then sum out c which eliminates all our factors of f(C) and f(ec) leaving us with f(e) where p(e) = 0.7(0.376) + 0.2(0.624) = 0.388