Your first calculation finds the probability that the person hits the target $4$ times in a row. That is very different (and much smaller) than the probability that the person hits at least once.
Let's do the problem in another way, much too long, but it will tell us what is going on.
What is the probability of hitting at least once? The event "at least one hit" could happen in several ways: (i) exactly $1$ hit; (ii) exactly $2$ hits; (iii) exactly $3$ hits; (iv) exactly $4$ hits.
(i) The probability of exactly one hit is $\binom{4}{1}(1/4)(3/4)^3$. This is because the hit could happen in any one of $4$ (that is, $\binom{4}{1}$) places. Write H for hit and M for miss. The probability of the pattern HMMM is $(1/4)(3/4)(3/4)(3/4)$. Similarly, the probability of MHMM is $(3/4)(1/4)(3/4)(3/4)$. You will notice this probability is the same as the probability of HMMM. We get the same probability for MMHM and for MMMH, for our total of $\binom{4}{1}(1/4)(3/4)^3$.
(ii) Similarly, the probability of exactly $2$ hits is $\binom{4}{2}(1/4)^2(3/4)^2$.
(iii) The probability of $3$ hits is $\binom{4}{3}(1/4)^3(3/4)$.
(iv) The probability of $4$ hits is $\binom{4}{4}(1/4)^4$. This is the $(1/4)^4$ that you calculated.
Add up. We get the required answer.
However, that approach is a lot of work. It is much easier to find the probability of no hits, which is the probability of getting MMMM. This is $(3/4)^4$. So the probability that the event "at least one hit" doesn't happen is $(3/4)^4$. So the probability that the event "at least one hit" does happen is $1-(3/4)^4$.
The probability that he hits the target on this 11th attempt is $0.8$ - you're assuming that past (independent, we assume) events have an effect on the outcome of the event.
On the other hand, the probability of that specific sequence of events, $10$ failures and $1$ success, is $$(0.2)^{10}(0.8) = 8.2\times10^{-8}$$
Best Answer
$$ \begin{align} P(\mbox{target is hit once}) &= P(\mbox{A hitting}) \cdot P(\mbox{B not hitting}) + P(\mbox{A not hitting}) \cdot P(\mbox{B hitting}) \\ &= \frac{1}{4}\cdot\frac{2}{3} + \frac{3}{4}\cdot\frac{1}{3} \\ &= \frac{5}{12} \end{align} $$
So, $$P(\mbox{A hitting | target is hit once}) = \frac{P(\mbox{A hitting}) \cdot P(\mbox{B not hitting})}{P(\mbox{target is hit once})} = \dfrac{\frac{1}{6}}{\frac{5}{12}} = \frac{2}{5}.$$