Heuristic reasoning (as well as numerical evidence) suggests that all of them are finite. Here is a discussion, with a couple of nice graphs. Here is a more focused article.
It looks like 12 is the largest even integer that can be expressed in only one way as a sum of two primes. Checking this up to some largish number (say, 1000000), and finding the corresponding maxima of your other subsets over that range, would make a nice programming exercise, which I might try when I have time.
I modified my program and have a few more results.
Unless my program has a bug there are no solutions for $n = 5, a,b,c \leq 4000$ or $n = 4, a,b,c \leq 10000$
For $n = 3$: There are 180 solutions for $a < b < c \leq 10000$.
With regards to the theory about sharing prime factors: There are cases where none, 2, or 3 of $a,b,c$ share at least one prime factor.
Additionally, there are some cases with $a < b < c$ and some with $a < c < b$.
This is a non-comprehensive list of my results:
$$\phi(16^3)+\phi(46^3)=\phi(45^3)$$
$$\phi(33^3)+\phi(231^3)=\phi(242^3)$$
$$\phi(90^3)+\phi(222^3)=\phi(228^3)$$
$$\phi(107^3)+\phi(354^3)=\phi(251^3)$$
$$\phi(360^3)+\phi(888^3)=\phi(912^3)$$
$$\phi(527^3)+\phi(700^3)=\phi(723^3)$$
$$\phi(1530^3)+\phi(1692^3)=\phi(1956^3)$$
For $n = 2$ where $a,b,c$ are not a Pythagorean triple:
There are a 547 solutions with $a,b,c \leq 200$
With $a < b$ there was actually at least one example for every value of a from $a = 1$ to $a = 100$.
Best Answer
There is an infinite number of such pythagorean triples. Any primitive pythagorean triple $(a,b,c)$ with $a^2+b^2=c^2$ (are we are looking for primitive triples since we want $a$ and $b$ consecutive) is of the form: $$ a=p^2-q^2,\qquad b=2pq,\qquad c=p^2+q^2 $$ with $p$ and $q$ coprime and not both odd. So we are looking for integer solutions of: $$ p^2-2pq-q^2 =\pm 1,$$ or: $$ (p-q)^2 - 2q^2 = \pm 1.$$ However we know that the Pell equation $A^2-nB^2=1$ has an infinite number of integer solutions $(A,B)$ for every $n$ that is not a square, hence we can find "consecutive" pythagorean triples from the solutions of $$ A^2 - 2B^2 = 1,\tag{1}$$ for istance. $(A,B)=(3,2)$ is the minimal solution of $(1)$, giving $(p,q)=(5,2)$, hence the triple $(20,21,29)$. The next solution can be found by expanding: $$ (3+2\sqrt{2})^2 = 17+12\sqrt{12},$$ hence $(p,q)=(29,12)$ gives the triple $(696,697,985)$ and so on.
In general, we can see that all the solutions depends on the convergents of the continued fraction of $\sqrt{2}$, i.e. on the Pell sequence: $$ (p,q) = (P_n,P_{n+1}),$$ from which:
where:
$$P_n = \frac{1}{2\sqrt{2}}\left((1+\sqrt{2})^n-(1-\sqrt{2})^n\right).$$
Since neither $2P_n P_{n+1}$ or $P_{n+1}^2-P_{n}^2$ may be primes if $n> 1$, the only "consecutive" pythagorean triple with the smallest element being a prime is $(3,4,5)$.