1) it's "any old element". the trick is to pick the right choices for a,b,c and d, so that we wind up with two products agb and agd we KNOW are equal. there are different ways to do this, here is one:

choose a = x, b = y, c = y, d = x, and g = x^{-1}

then agb = x(x^{-1})y = (xx^{-1})y = y, and:

cgd = y(x^{-1})x = y(xx^{-1}) = y, so we have agb = cgd. because in G agd = cgd implies ab = cd, we have: xy = yx, so G is abelian.

2) hint: if no element except the identity is its own inverse, multiply all the elements of G together like so:

(e)(g_{1}g_{1}^{-1})(g_{2}g_{2}^{-1})....... = ?

(of course this means that |G| must be odd. why is this not a problem?)