3 clarifications due to imprecission. sorry for the editng noise

I agree that the answer to the first question should be more or less nothing. However, I disagree that this is for 'the same types of issues surrounding the ABC-conjecture', and so I highly doubt there is much of a direct link.

The ABC-conjecture is very different in that (in a certain sense) it says something on the actual divisors of the involved numbers, not merely on their number.

The reason why in my opinion the answer to the first question is essentially nothing is that simply there should be little meaninful to say, more or less everything should be possible at least on a quantitative level (that is not obviously impossible)relating the respective approx. sizes).

To wit take $b=1$. If $a$ is a Sophie Germain prime then $a+b$ will only have $4$ divisors, but it is not at all true that $a+1$ has few prime divisors for any $a$ prime (ie any $a$ with $\tau(a)=2$).

Or, it is kown that in some sense there is 'nothing special' regarding divisors counts (except for evenness issues) of $p-1$ for $p$ prime relative to $n-1$ for $n$ any number.

(See also this MO question http://mathoverflow.net/questions/27508/factors-of-p-1-when-p-is-prime ; and I believ, but only checked briefly, the paper mentioned in a comment also goes roughly speaking in this direction; the result on something relating $\tau(n)$ and $\tau(n+a)$ coinciding with the prediction one has if considering the two 'independently').

And $b=1$ is not really special in that regard.

Or it is known that many (and if Goldbach conj. is true, all except one) even number can be written as sum of two primes, so (essentially) whatever even $n$ you consider you can write it as $n=a+b$ with $\tau(a)=\tau(b)=2$, so $\tau(a)=\tau(b)=2$ implies really (next to) nothing on $\tau(a+b)$. (And if you impose larger values for $\tau(a)$ and $\tau(b)$ it is rather simmpler, as there will be more such numbers, to show that you can write (essentially) whatever $n$ as a sum of two such $a,b$.)

So, in brief, yes there should be very little to be said (as regards to connecting these values), yet no this is not much related to ABC, in particular it is not that things are not known because they are diffcult but there is actaullt actually hardly any connection between $\tau(a+b)$ and $\tau(a)$, $\tau(b)$.\tau(b)$on a quantitative level. (If one were to investigate specific restrictions on the$\tau$, in particular such that they force/are equivalent too the number being a high power then the type of questions become a bit different and then there is I think some link to ABC; but then to me they are not really questions on$\tau$but 'something else' encoded via$\tau$.) 2 added additional reasoning I agree that the answer to the first question should be more or less nothing. However, I disagree that this is for 'the same types of issues surrounding the ABC-conjecture', and so I highly doubt there is much of a direct link. The ABC-conjecture is very different in that (in a certain sense) it says something on the actual divisors of the involved numbers, not merely on their number. The reason why in my opinion the answer to the first question is essentially nothing is that simply there should be little meaninful to say, more or less everything should be possible (that is not obviously impossible). To wit take$b=1$. If$a$is a Sophie Germain prime then$a+b$will only have$4$divisors, but it is not at all true that$a+1$has few prime divisors for any$a$prime (ie any$a$with$\tau(a)=2$). Or, it is kown that in some sense there is 'nothing special' regarding divisors counts (except for evenness issues) of$p-1$for$p$prime relative to$n-1$for$n$any number. (See also this MO question http://mathoverflow.net/questions/27508/factors-of-p-1-when-p-is-prime ; and I believ, but only checked briefly, the paper mentioned in a comment also goes roughly speaking in this direction; the result on something relating$\tau(n)$and$\tau(n+a)$coinciding with the prediction one has if considering the two 'independently'). And$b=1$is not really special in that regard. Or it is known that many (and if Goldbach conj. is true, all except one) even number can be written as sum of two primes, so (essentially) whatever even$n$you consider you can write it as$n=a+b$with$\tau(a)=\tau(b)=2$, so$\tau(a)=\tau(b)=2$implies really (next to) nothing on$\tau(a+b)$. (And if you impose larger values for$\tau(a)$and$\tau(b)$it is rather simmpler, as there will be more such numbers, to show that you can write (essentially) whatever$n$as a sum of two such$a,b$.) So, in brief, yes there should be very little to be said (as regards to connecting these values), yet no this is not much related to ABC., in particular it is not that things are not known because they are diffcult but there is actaullt hardly any connection between$\tau(a+b)$and$\tau(a)$,$\tau(b)$. 1 I agree that the answer to the first question should be more or less nothing. However, I disagree that this is for 'the same types of issues surrounding the ABC-conjecture', and so I highly doubt there is much of a direct link. The ABC-conjecture is very different in that (in a certain sense) it says something on the actual divisors of the involved numbers, not merely on their number. The reason why in my opinion the answer to the first question is essentially nothing is that simply there should be little meaninful to say, more or less everything should be possible (that is not obviously impossible). To wit take$b=1$. If$a$is a Sophie Germain prime then$a+b$will only have$4$divisors, but it is not at all true that$a+1$has few prime divisors for any$a$prime (ie any$a$with$\tau(a)=2$). Or, it is kown that in some sense there is 'nothing special' regarding divisors counts (except for evenness issues) of$p-1$for$p$prime relative to$n-1$for$n$any number. (See also this MO question http://mathoverflow.net/questions/27508/factors-of-p-1-when-p-is-prime ; and I believ, but only checked briefly, the paper mentioned in a comment also goes roughly speaking in this direction; the result on something relating$\tau(n)$and$\tau(n+a)$coinciding with the prediction one has if considering the two 'independently'). And$b=1\$ is not really special in that regard.

So, in brief, yes there should be little to be said (as regards to connecting these values), yet no this is not much related to ABC.