I assume $P(x_i = x_j) = 0$ (e.g., when the $x_i$ are continuous random variables).
Each ordering of the six random variables are equiprobable (with probability $P(x_1 < x_2 < x_3 < x_4 < x_5 < x_6) = 1/6!$). So just count permutations.
$$P(x_6 > x_2 \mid x_1 = \max\{x_1, \ldots, x_5\})
= \frac{P(x_6 > x_2 ,x_1 = \max\{x_1, \ldots, x_5\})}{P(x_1 = \max\{x_1, \ldots, x_5\})}
= \frac{3!(2+3+4+5)/6!}{6 \cdot 4! / 6!} = \frac{7}{12}.$$
e.g., for the denominator, there are $4!$ ways to put $x_1, \ldots, x_5$ in order such that $x_1$ is the largest; then there are $6$ places to position $x_6$ anywhere amongst the others. For the numerator I had to do some casework based on where $x_2$ is positioned among $x_2, \ldots, x_5$.
When $P(x_i = x_j) \ne 0$ (e.g., when the $x_i$ have a discrete component), then we need more information to answer the question.