If you want to see why the sum converges to 1/3, you can write out the power series. It basically amounts to the fact that mu*1=delta, where mu is the mobius function, * represents Dirichlet convolution, and delta(n)=1 if n=1 and 0 otherwise.

The main idea (for the second one) is that we have that the sum over the positive integers of mu(n)/(3^(n)-1)=1/3. We think about the following range: (1/3-1/(2*3^(n)-2)), 1/3+1/(2*3^(n)-2))). I claim that for any sum over n>m of (a_n)/(3^n-1) with a_n=-1,0, or 1 to converge to x, |x| must be less than 1/(2*3^(m)-2). We can see this using geometric series (looking at the sum of the terms with n>m and assuming a_n=1 for all n). However, there is always only one choice of a_n that puts the partial sum in (1/3-1/(2*3^(n)-2)), 1/3+1/(2*3^(n)-2))). This is how it works!