Well if 3*33 = -1 (mod 100), then it stands to reason that:

3*(-33) = 1 (mod 100), that is, the inverse of 3 (mod 100) is -33 = 67.

If we cube this, we get:

(67)^{3}= (-33)^{3}= (-33)(33)^{2}= (-33)(3^{2})(11^{2}) = (-33)(9)(121) = (-33)(3)(3)(121) = (3)(121) = (3)(21) = 63 (mod 100).

Your second question is harder to answer.

Assuming the answer IS 3^{-9}mod 100, we calculate:

3^{-9}= (3^{-1})^{9}= 67^{9}= [(67)^{2}]^{4}(67) = (89)^{4}(67) = (-11)^{4}(67) = (11)^{4}(67) = (21)^{2}(67) = (41)(67) = 47.