p(r1)=1/3p(r2)
p(r1)+p(r2)=4p(r2)=14/25
p(r2)=14/100
1-p(r2)=1-14/100=86/100

Careful with your calculations - if it is raining on only one of the two days, then we have to consider that it could rain on Day 1 but NOT Day 2, or it could NOT rain on Day 1 but rain on Day 2.

In the absence of any complete explanation to the DS, I decided to show my way to attempt this hard DS. A DS can be termed to be hard if it forces a test taker to finish calculations before making a decision about it, which doesn't happen so frequently on GMAT. Talaangoshtari's approach is losing its way as it's definitely not interpreting statement 1 to it's real meanings, which Dave has fittingly mentioned in his short post.

I'd prefer to resort to some Algebra here, by taking x as the probability that it'll rain on Day 1, so that 3x is the probability that it'll rain on Day 2. Therefore, probability that it won't rain on Day 1 of the tournament is (1 - x) and probability that it won't rain on Day 2 of the tournament is (1 - 3x). In short, we need to answer:

(1 - 3x) = ? Or just answer x = ?

(1) The right interpretation of this statement is:

Probability (D1 Yes & D2 No OR D1 No & D2 Yes) = 14/25

Or, x(1 - 3x) + (1 - x).3x = 14/25

It's hard to make any decision from here, rare case on GMAT, we need to continue and be our number savvy:

We get x - 3x^2 + 3x - 3x^2 = 14/25

Or, 4x - 6x^2 = 14/25

Or, 2x - 3x^2 = 7/25, and finally this quadratic at hand:

75x^2 - 50x + 7 = 0, get your bearings champ!

75x^2 - 15x - 35x + 7 = 0

(5x - 1)(15x - 7) = 0

Hence, x = 1/5 or x = 7/15. Hold on! Please do not call this statement insufficient, because it's a hard GMAT DS. Just think over, x = 1/5 is fair enough, but, can x equal 7/15 too (?); imagine, if x equal 7/15, then 3x equals 7/5. Impossible possibility, no! Hence, only x = 1/5 is acceptable. [spoiler]Sufficient

BCE gone, AD on.[/spoiler]

(2) Isn't it a useless statement? Something we already agree to. A useless statement is always [spoiler]insufficient.

Hence A[/spoiler]