Actually, that's not how the probability works.
With the first example, the probabilities you have to add up are those of hitting with the first and missing with the second, missing with the first and hitting with the second, and hitting with both. This comes out to (0.5*0.75)+(0.5*0.25)+(0.5*0.25) or 0.625, 62.5%.
With the second example, it's more complex. The outcomes and their probabilities are (H = hit, M = miss. I'm a lazy typist):
HMM, 0.204
MHM, 0.204
HHM, 0.136
HMH, 0.036
MHH, 0.036
MMH, 0.054
HHH, 0.024
This comes out to a probability of a hit in the round being 0.694, or 69.4%.
So yes, the chance of scoring a hit is increased marginally, but it's nowhere near 95%.
Obviously for you other probability people out there, if I have made a mistake or missed out an outcome, please feel free to correct me. I'm as fallible as the next man, if not moreso.
EDIT: Of course, the easier way of doing it is working out the chance of missing everything and taking it away from 100%.
It the first example, this is (0.5*0.75) = 0.375, which, when taken away from 1 leaves 0.625, which is the same as above.
In the second example, it's (0.6*0.6*0.85) = 0.306, which, when taken away from 1, leaves 0.694, which is the same as above. I can't believe it took rereading the post to realise this...