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Let $p$ be the probability of hitting the target and $q$ be the probability of missing the target. Given $p = 3q$. Since $p + q = 1$, we have $3q + q = 1$, which implies $4q = 1$. Thus, $q = 0.25$ and $p = 0.75$ (or $3/4$).
The sniper fires two shots. The probability of missing both shots is $q \times q = (1/4) \times (1/4) = 1/16$. The probability that the target is hit at least once is $1 - P(\text{missing both}) = 1 - 1/16 = 15/16$.
The probability that at least one shot misses is the complement of the event that both shots hit the target. $P(\text{both hit}) = p \times p = (3/4) \times (3/4) = 9/16$. Therefore, $P(\text{at least one miss}) = 1 - 9/16 = 7/16$.
Final Answer: (i) 15/16, (ii) 7/16