The number of common terms in the progressions $$4, 9, 14, 19, \ldots$$ up to $$25^{th}$$ term and $$3, 6, 9, 12, \ldots$$ up to $$37^{th}$$ term is :

First AP: $$4, 9, 14, 19, \ldots$$ with $$a_1 = 4, d_1 = 5$$.

25th term: $$a_{25} = 4 + 24 \times 5 = 124$$.

General term: $$a_n = 4 + 5(n-1) = 5n - 1$$. Terms: $$4, 9, 14, ..., 124$$.

Second AP: $$3, 6, 9, 12, \ldots$$ with $$a_1 = 3, d_2 = 3$$.

37th term: $$a_{37} = 3 + 36 \times 3 = 111$$.

General term: $$b_m = 3m$$. Terms: $$3, 6, 9, ..., 111$$.

Common terms satisfy: $$5n - 1 = 3m$$, i.e., $$5n - 1 \equiv 0 \pmod{3}$$, so $$5n \equiv 1 \pmod{3}$$, so $$2n \equiv 1 \pmod{3}$$, so $$n \equiv 2 \pmod{3}$$.

So $$n = 2, 5, 8, 11, 14, 17, 20, 23, \ldots$$ (up to 25).

The common terms are $$5n - 1$$ for these values of $$n$$: $$9, 24, 39, 54, 69, 84, 99, 114$$.

Now check which are also $$\leq 111$$ (upper bound of second AP):

$$9, 24, 39, 54, 69, 84, 99$$ are all $$\leq 111$$. ✓

$$114 > 111$$. ✗

So there are 7 common terms.

The answer is $$\boxed{7}$$, which corresponds to Option (3).