What is the probability of rolling 2 7's before 6 even numbers on successive rolls of a pair of fair dice? Would this equal Pr( rolling 2 7's)/Pr(rolling 2 7's or rolling 6 even numbers)? Or could I approach the problem as follows:
1- (Pr(rolling 6 even numbers)+ Pr( rolling 5 even numbers, 7 and an even number)) ?

Here’s a slightly different way to look at it.
Odd rolls other than seven are meaningless and can be ignored. By the seventh meaningful roll you must have either two sevens or six even numbers. Thus, you get two sevens before you get six even numbers if and only if you get at least two sevens in the first seven rolls.
On each meaningful roll the probability of getting a seven is $$\frac{\frac16}{\frac16+\frac12} = \frac14,$$ so the probability of getting at most one seven in seven meaningful rolls is $$\binom70\left(\frac34\right)^7 + \binom71\left(\frac14\right)\left(\frac34\right)^6 = \frac{3^7+7\cdot 3^6}{4^7} = \frac{10\cdot 3^6}{4^7} = \frac{3645}{8192}.$$ 
The probability of getting at least two sevens in seven meaningful rolls is therefore $$1 - \frac{3645}{8192} = \frac{4547}{8192}.$$
