For the first if either diverges the whole thing diverges
So yopu only have to consider 3/(n+1)
You could of course do a comparison
(3n+2)/(n^2 +n) > 3n/(n^2 +n) = 3/(n+1) > 3/(2n)= (3/2)(1/n)
which is your divergent harmonic sereies
For #2 is ln(n)^2 in the numerator ?
you can always work out the AD first and use the original limits
but still no problem as you'll end up with [ (ln(ln(t))]^3 which -> infinity