As Ross has pointed out, the parity of the statement is known beforehand in cases $2$ to $4$, and the value of $X$ is irrelevant, so there's no information gain in these cases.
In case $1$, the probability of having a certain rank in a uniformly random poker hand is $1-(1-1/13)^5=122461/371293$. The probability of having a certain rank given a flush is $5/13$. The information you gain if he has the rank is measured by the Kullback–Leibler divergence
$ P(\text{flush}\mid X)\log\frac{P(\text{flush}\mid X)}{P(\text{flush})}+P(\overline{\text{flush}}\mid X)\log\frac{P(\overline{\text{flush}}\mid X)}{P(\overline{\text{flush}})}\;. $
With
$ P(\text{flush})=\frac{4\binom{13}5}{\binom{52}5}=\frac{33}{16660}\approx0.00198 $
and
$ \frac{P(\text{flush}\mid X)}{P(\text{flush})}=\frac{P(X\mid\text{flush})}{P(X)}=\frac{5\cdot371293}{13\cdot122461}=\frac{142805}{122461}\approx1.166 $
and
$ \frac{P(\overline{\text{flush}}\mid X)}{P(\overline{\text{flush}})}=\frac{1-P(\text{flush}\mid X)}{1-P(\text{flush})}=\frac{16660-33\cdot142805/122461}{16660-33}=\frac{2035487695}{2036159047}\approx0.99967\;, $
the divergence is
$ \Delta_X=\frac{33}{16660}\cdot\frac{142805}{122461}\log\frac{142805}{122461}+\left(1-\frac{33}{16660}\cdot\frac{142805}{122461}\right)\log\frac{2035487695}{2036159047}\approx0.000025988 $
(computation). If he doesn't have the rank, the divergence is
$ P(\text{flush}\mid\overline X)\log\frac{P(\text{flush}\mid\overline X)}{P(\text{flush})}+P(\overline{\text{flush}}\mid\overline X)\log\frac{P(\overline{\text{flush}}\mid\overline X)}{P(\overline{\text{flush}})}\;. $
With
$ \frac{P(\text{flush}\mid\overline X)}{P(\text{flush})}=\frac{P(\overline X\mid\text{flush})}{P(\overline X)}=\frac{8\cdot371293}{13\cdot248832}=\frac{28561}{31104}\approx0.918 $
and
$ \frac{P(\overline{\text{flush}}\mid\overline X)}{P(\overline{\text{flush}})}=\frac{1-P(\text{flush}\mid\overline X)}{1-P(\text{flush})}=\frac{16660-33\cdot28561/31104}{16660-33}=\frac{172416709}{172388736}\approx1.00016\;, $
the divergence is
$ \Delta_{\overline X}=\frac{33}{16660}\cdot\frac{28561}{31104}\log\frac{28561}{31104}+\left(1-\frac{33}{16660}\cdot\frac{28561}{31104}\right)\log\frac{172416709}{172388736}\approx0.0000068215 $
(computation). The expected information gain is therefore
$ P(X)\Delta_X+P(\overline X)\Delta_{\overline X}\approx0.00001314\;, $
or roughly $0.000019$ bits.