For 3, that's a great start! So 9 must be a factor of 14b^2, meaning actually 3 has to be a factor of b. So b = 3*y for some y. Try substituting this back into your fraction (9a^2+14b^2)/(9ab). You should be able to simplify a little now proceed somewhat similarly to the start. Remember that gcd(a,b) = 1!

9 is quite challenging. There are some fancier tricks but I think it's best to start just with some basics. Note any three-digit N has at most 5 digits written in base 5 (since 5^5 > 1000) and at most 4 digits in base 6 (since 6^4 > 1000). Hence we can write$$N = 625a + 125b + 25c + 5d + e = 216w + 36x + 6y + z$$for a,b,c,d,e base-5 digits (so less than 5) and w, x, y, z base-6 digits (so less than 6).

From here, we know (from the sum as "base-10 numbers")$$2N \equiv 10d + e + 10y + z.\pmod{100}.$$Note, for example, we can write $2N = 1250a + 250b + 50c + 10d + 2e$. Can we use this to start getting some restrictions on the digits? As a hint, you should be able to start by showing that e and z must be equal.