A) Write the two half reactions:
Au3+ + 3e- --> Au, Eo = +1.5V
Li+ + 1e- --> Li, Eo = -3.05
Which half reaction has a higher reduction potential? The first one.
So, the 2nd reaction must be the one that is oxidized. So, let's flip the reaction and the sign of Eo for this reaction:
Au3+ + 3e- --> Au, Eo = +1.5V
Li --> Li+ + 1e-, Eo = +3.05
B) Multiply the two reaction with proper coefficients, so that you can cancel out the electrons (DO NOT multiply the reduction potential by these coefficients):
1(Au3+ + 3e- --> Au), Eo = +1.5V
3(Li --> Li+ + 1e-), Eo = +3.05
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Au3+ + 3Li --> Au + 3Li+, Eo = 1.5 + 3.05 = 4.55V