Binomial Theorem

For Federer to win on the 10th point, he has to win exactly 6 out of the 9 previous points.
By binomial theorem, he can do this with probability P= $$^9 C _6 * (0.7)^6 * (0.3)^{3} = 84 * 0.7^6 * 0.3^3$$.
The probability that he wins the 10th point is 0.7* P i.e. $$84 * 0.7^7 * 0.3^3$$.

The probability p of hitting the target is 1/5.
Hence, the probability of missing the target on all 10 shots = $$(0.8)^{10}$$.
The probability of hitting the target at least twice = 1- missing the target - hitting the target exactly once.

Probability of hitting exactly once = $$^{10} C _1 * (0.2) * (0.8)^{9} = 2.5 * (0.8)^{10}$$.

Hence, the probability of hitting at least twice = $$1- (0.8)^{10} - 2.5* (0.8)^{10} = 1 - 3.5*(0.8)^{10}$$.

Let the term with coefficient $$x^8$$ in the expansion $$(px^2+\frac{1}{qx})^{13}$$ be the $$a^{th}$$ term.
So, 2*a-(13-a) = 8
or a=7
So, the coefficient of $$x^8$$ is $$^{13} C_7 *p^7*q^{-6}$$

Similarly, the term with coefficient $$x^{-8}$$ in the expansion $$(px-\frac{1}{qx^2})^{13}$$ is the sixth term.
So, the coefficient of $$x^{-8}$$ is -$$^{13} C_6 * p^6*q^{-7}$$

As, both of them are equal, pq = -1