6 A continuous random variable \(X\) takes values from 0 to 6 only and has a probability distribution that is symmetrical. Two values, \(a\) and \(b\), of \(X\) are such that \(\mathrm { P } ( a < X < b ) = p\) and \(\mathrm { P } ( b < X < 3 ) = \frac { 13 } { 10 } p\), where \(p\) is a positive constant.

1. Show that \(p \leqslant \frac { 5 } { 23 }\).
2. Find \(\mathrm { P } ( b < X < 6 - a )\) in terms of \(p\).
   It is now given that the probability density function of \(X\) is f , where $$f ( x ) = \begin{cases} \frac { 1 } { 36 } \left( 6 x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 6
   0 & \text { otherwise } \end{cases}$$
3. Given that \(b = 2\) and \(p = \frac { 5 } { 27 }\), find the value of \(a\).