I'll only answer the first question. The probability that someone has a better flush can be calculated with similar methods, but the calculation is considerably more complicated.
I presume that you don't mean an actual game of Texas hold 'em, in which there would have been betting and folding at earlier stages, which would render the assumption of uniform distribution of hands highly unrealistic; so this is about a fictitious game in which five cards are dealt to the table and two cards are dealt to each hand, with no intermediate betting.
There are three possibilities: There might be $3$, $4$ or $5$ cards of the same suit on the table.
The probability for $5$ cards of the same suit on the table is
$4\frac{\binom{13}5}{\binom{52}5}=\frac{33}{16660}\approx0.002\;.$
In this case everyone has a flush.
The probability for exactly $4$ cards of the same suit on the table is
$4\frac{\binom{13}4\binom{39}1}{\binom{52}5}=\frac{143}{3332}\approx0.043\;.$
In this case, your probability of having one card of the suit is
$\frac{\binom91\binom{38}1}{\binom{47}2}=\frac{342}{1081}\approx0.316\;,$
and
$\frac{\binom92}{\binom{47}2}=\frac{36}{1081}\approx0.033$
for two cards of the suit. If you have one card, the probability of someone else having at least one is
$ 1-\frac{\binom{37}6}{\binom{45}6}=\frac{6299}{8815}\approx0.715\;, $
whereas if you have two cards, it is
$ 1-\frac{\binom{38}6}{\binom{45}6}=\frac{23309}{35260}\approx0.661\;. $
The probability for exactly $3$ cards of the same suit on the table is
$4\frac{\binom{13}3\binom{39}2}{\binom{52}5}=\frac{2717}{8330}\approx0.326\;.$
In this case your probability of having two cards of the suit is
$\frac{\binom{10}2}{\binom{47}2}=\frac{45}{1081}\approx0.042\;.$
The probability of someone else then also having two cards of the suit is a bit more complicated to calculate. We can get it by applying the inclusion–exclusion principle:
$3\frac{\binom82}{\binom{45}2}-3\frac{\binom84}{\binom{45}4}+\frac{\binom86}{\binom{45}6}=\frac{8091}{96965}\approx0.083\;.$
To put this all together, we have to divide the probability of you and someone else having a flush by the probability of you having a flush:
$\frac{\frac{33}{16660}+\frac{143}{3332}\left(\frac{342}{1081}\cdot\frac{6299}{8815}+\frac{36}{1081}\cdot\frac{23309}{35260}\right)+\frac{2717}{8330}\cdot\frac{45}{1081}\cdot\frac{8091}{96965}}{\frac{33}{16660}+\frac{143}{3332}\left(\frac{342}{1081}+\frac{36}{1081}\right)+\frac{2717}{8330}\cdot\frac{45}{1081}}=\frac{145640554}{323494633}\approx0.450$
(computation). Thus the chance of someone having a flush if you have one is about $45\%$. This is confirmed by computer simulations.