Could someone help me clear up some confusion I have over independence/dependence of R.V :
Question 1) Say we have two random variables $X$ and $Y,$ with $Y$ dependent on $X$
$X$ distributed $\operatorname{Uniform}(0,1).$
$Y$ distributed $$ F_Y(y) = \begin{cases} 0, & y<0 \\ 1, & y>1 \\ y, & y\in [0,1] \end{cases} $$ i.e it is $\operatorname{Uniform}(0,1),$ however let $Y=\frac{1}{2}$ if $X\in (0,\frac{1}{2})$ so $Y$ depends on $X.$
Then am I right in saying Y is still distributed $\operatorname{Uniform}(0,1)$ since its C.D.F is unchanged?
Question 2)
Say $X$ and $Y$ are distributed as Bernouli R.V i.e they are $1$ with probability $p$ and $0$ with probability $1-p.$
However it is also the case that if $X=1$ then $Y=0$ and if $Y=0$ then $X=1$ so they are dependent on each-other.
Then how would one calculate $P(X=1,Y=0)$ since there is no "conditioned on" does this mean (in a real world sense) we are running both experiments simultaneously and so neither has a dependence on the other?
The distribution of $X$ implies that $\Pr(X<1/2) = 1/2,$ so $\Pr(Y=1/2) \ge 1/2,$ and thus the c.d.f. of $Y$ is not what you reported it to be; $Y$ is not uniformly distributed.
In your case of Bernoulli distributions, you say $\Pr(X=1) = \Pr(Y=1) = p$ and $\Pr(X=0) = \Pr(Y=0) = 1-p.$ And you also say $X=0$ if, but only if, $Y=1.$ That implies $$ 1-p = \Pr(X=0) = \Pr(Y=1) = p. $$ Thus $1-p=p,$ so $p=1/2.$
You have $\Pr(X=1\ \&\ Y=0) = 1/2$ and $\Pr(X=0\ \&\ Y=1) = 1/2.$
You can also say $\Pr(X=1\ \&\ Y=0) = \Pr(X=1)\cdot\Pr(Y=0\mid X=1) = \dfrac 1 2 \cdot 1.$