Distinguishability is not a property of the balls. It is the property of the observer. If there are balls of different colors arranged in boxes then certain arrangements will be distinguishable for those with a good eyesight and some arrangements will not be distinguishable for the colorblind.
In probability theory objects are called indistiguishable if the observer (experimenter) finds those arrangements equally likely that he can distinguish when, in theory, other observers with better eyes could distinguish further arrangements.
That is: "indistiguishability" is a misnomer.
So your answer is correct whether or not you are able to see the numbers on the balls. If you are then you would have more elementary events but you would have to unite some.
See again André Nocolas' answer.
There are several approaches to this problem. One of the simplest to explain is that you can make a distribution chart to describe the scenario:
$$\begin{array}{c|c|c|c|c|c|c|}
& \color{blue}{1} & \color{red}{1} & \color{green}{1} & \color{yellow}{1}&\color{blue}{2} & \color{red}{2}&\cdots\\
\hline
\color{blue}{1} & 0 & x & x & \cdots\\
\hline
\color{red}{1} & x & 0 & x & \cdots\\
\hline
\color{green}{1} & x & x & 0 & x & \cdots\\
\hline
\color{yellow}{1}&\vdots & \ddots & \ddots & \ddots\\
\hline
\color{blue}{2} \\
\hline
\color{red}{2}\\
\hline
\vdots\\
\end{array}$$
(note: these probabilities have it such that order matters. The final answer in this problem will not depend on whether you use a method where order matters or doesn't matter, so use what is comfortable)
Noting that it is impossible to draw the same card twice in a row (hence the main diagonal being zeroes) and all other entries are equiprobable and should add up to 1, you calculate $Pr(\text{first card is}~ a\cap \text{second card is}~ b) = \frac{1}{16\cdot 15}$. You may then note that there are $48$ squares with nonzero probability that correspond to having the same number in both the first draw and the second draw. Use addition principle to complete the argument for part (a). For part (b), add up the probability of all squares whose corresponding outcomes have the total sum is bigger than 6.
That approach is rather tedious and would require either drawing a very large chart, or only drawing a portion of it (like I did) and "reading" parts of the chart that you haven't written down yet. Instead, let us consider this via multiplication and addition principles.
$Pr(\text{first two numbers match}) = Pr(\text{both numbers are 1}\cup \text{both numbers are 2}\cup\text{both numbers are 3}\cup\text{both numbers are 4})\\
=Pr(\text{both numbers are 1}) + Pr(\text{both numbers are 2}) + Pr(\text{both numbers are 3}) + Pr(\text{both numbers are 4})$
Here, I was able to split up the unions ($\cup$) by the addition principle:
If $A\cap B=\emptyset$ then $Pr(A\cup B) = Pr(A)+Pr(B)$
More generally, for any $A$ and $B$ you have $Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$
Now, we wish to solve for $Pr(\text{both numbers are 1})$. This is the same as $Pr(\text{first is a 1}\cap \text{second is a 1})$. To do this, we use the multiplication rule:
$Pr(A\cap B) = Pr(A)\cdot Pr(B|A)$
So, we look at what $Pr(\text{first is a 1})$ is and what $Pr(\text{second is a 1}|\text{first is a 1})$. The probability the first card is a 1 is simply $\frac{4}{16}$ (since there are four 1's out of sixteen total cards) and the probability of the second being a 1 given that the first is a one as well is $\frac{3}{15}$ (since there will be three remaining 1's out of fifteen remaining cards total).
Thus, we see $Pr(\text{both are 1}) = \frac{4}{16}\cdot\frac{3}{15}$. Through a similar argument we find that the rest of the probabilities are also $\frac{4}{16}\cdot\frac{3}{15}$ as well. So, $Pr(\text{both numbers the same}) = 4\cdot Pr(\text{both are 1}) = 4\cdot \frac{4}{16}\cdot \frac{3}{15} = \frac{3}{15}$
Once you become more comfortable with calculating probabilities, you may make simplifications which save a great deal more time. As mentioned as a hint in the comments, it doesn't actually matter what the first number drawn is. As such $Pr(\text{both are same number}) = Pr(\text{second number is same as first}) = \frac{3}{15}$ since whatever number the first card was, there will be three remaining of that number out of 15 cards to take from.
For part (b), use the tools described here and note that $Pr(\text{sum of cards is more than 6}) = Pr(\text{sum of cards is 7}\cup \text{sum of cards is 8})$. Note further that you can break $Pr(\text{sum of cards is 7})$ up as $Pr(\text{first card is 3 and second is 4}\cup\text{first card is 4 and second is 3})$.
Best Answer
This answer does not give an answer to the general question for large $N$, but at least shows a method by which you can compute the optimal strategy for any specific $N$. This is achieved by formulating the problem as finding a specific matching on a bipartite graph.
Let each node on one side, lets say $A$, represent the combination of cards you can draw and the nodes on the other side, $B$, represent the possible choices your friend can see. An edge in this graph will connect $a \in A$ with $b \in B$ iff $a$ can become $b$ by removing a card. A good strategy for this game would consist of a matching on the graph. Why? Because when you get a combination of cards, you have to have made a choice of which card to remove or which node to pair it up with. The matching also tells your friend what to guess: he should just guess the combination that his combination is paired up with.
But what matching corresponds to the optimal strategy? Well, we want to maximize the number of times we'll get a combination in the matching so if $M$ is the set of all matched nodes in $A$, and $w(a)$ is the chance of getting $a$, we want to maximize $$ \sum_{a \in M} w(a) $$
This corresponds to a maximum vertex-weighted matching problem which is a special case of the Assignment Problem. I made a program to solve this problem in C (source can be found here). The optimal strategies will have a success rate equal to the above sum. This will be a fraction where the denominator will be the number of combinations you can get, $N^2 \choose N$ (see A014062). The numerator does not seem to follow any clear pattern. I've calculated the numerator for some small values of $N$ with the program. These are tabulated below.
\begin{array}{ |r|r|r|r| } \hline N & \text{numerator} & \text{denominator} & \text{success rate} \\ \hline 2 & 5 & 6 & 0.833 \\ 3 & 72 & 84 & 0.857 \\ 4 & 1580 & 1820 & 0.868 \\ 5 & 45250 & 53130 & 0.852 \\ 6 & 1705347 & 1947792 & 0.876 \\ 7 & 73502156 & 85900584 & 0.856 \\ 8 & 3828899328 & 4426165368 & 0.865 \\ 9 & 225664027650 & 260887834350 & 0.865 \\ 10 & 14914814271500 & 17310309456440 & 0.862 \\ 11 & 1114114517120572 & 1276749965026536 & 0.873 \\ 12 & 89759340344067264 & 103619293824707388 & 0.866 \\ 13 & 7983517393715262672 & 9176358300744339432 & 0.870 \\ \hline \end{array}
Because of the exponential nature of this problem, calculating the success rate for larger $N$ would be pretty hard – at least with this approach.