For $1\over 3$, you get $0.\overline3$, which is $0.33333...$. The threes go on forever. You can't ask "What happens if it ends in an eight?" because it simply doesn't end.
For SSSSS..., what if it ends in a T? Well, an infinite series of Ss followed by a T would never have a T because it never ends.
There can be infinite points in a point, even a very small one.
Are they true or false? I think they're true, aren't they? If not, why not?
One thing you could do is to consider this as a limit. Note first that:
$$33333=\frac {10^5-1}3$$
Then $$333338=10\frac {10^5-1}3+8=\frac{10^6-10+24}3=\frac {10^6+14}3$$
Then $$0.333338=10^{-6}\cdot\frac {10^6+14}3=\frac {1+14\times 10^{-6}}3$$
Now if we have $r$ threes followed by an eight, the equivalent formula is $$\frac {1+14\times 10^{-(r+1)}}3=\frac 13+\frac {14}3\cdot10^{-(r+1)}$$
The value of this expression, as $r$ gets larger and larger, gets closer and closer to $\cfrac 13$ - we say that the limit is one third.
This may not be quite how you were thinking of the question - my daughter asks me questions like this sometimes, and I do my best to explain them. Really there isn't a last digit, and if you think about it you could add all sorts of noise at the end and still get the same limit.
But the insights which arise from your intuition helped mathematicians like Dedekind and Cauchy to define the real numbers and the meaning of limits, so that it made sense to have a single real number which was a limit of all kinds of different sequences in such a way that our decimal expansions of numbers still make sense.
