$$Y=[X+1]=i+1$$ iff $$i\le X \lt i+1,i=0,1,....$$
My approach to derive the distribution of $Y$ is to look at
$P(Y=y)=P(Y\le y)-P(Y\le y-1)$
Usually, I am ok to find $g^{-1}$ given $Y=g(X)$ but for this question, I am not sure how to find the inverse of $Y=[X+1]$.
So I looked at $Y=1 $, $P([X+1]=1)$ $\Rightarrow$ $P(0<X<1)$... And then extend it to when $Y=i$... which is $P(Y=i)=$so basically I am trying to find the pattern when solving for the inverse from $Y$ to $X$... What I found is that $P(Y\le i)=P(X<i)$.
I am wondering if there are other ways to solve the problem?
To continue:
$P(Y=y)=P(Y\le y)-P(Y \le y-1)=P(X<y)-P(X<y-1)=e^y-e^{-(y-1)}$ for $y=1,2,...$
The second question asks me to derive conditional distribution $P(X-4|y\geq 5)$ and I am completely lost.
My attempt is to use $$\frac{P(X-4,Y\geq5)}{P(Y\geq5)}$$ but I don't know what's the joint distribution of X and Y....
Concerning the first question: nothing is wrong with stating that:$$Y=n\iff n-1\leq X<n\text{ for }n=1,2,\dots$$so that also:$$P(Y=n)= P(n-1\leq X<n)\text{ for }n=1,2,\dots$$
Working this out leads to: $$P(Y=n)=P(X<n)-P(X<n-1)=(1-e^{-n})-(1-e^{-(n-1)})=e^{-n+1}-e^{-n}$$ for $n=1,2,\dots$
Note that $Y\geq5\iff X\geq4$ so that:$$P(X-4\geq x\mid Y\geq5)P(X\geq 4)=P(X-4\geq x\mid X\geq4)P(X\geq 4)=$$$$P(X-4\geq x, X\geq4)=P(X\geq\max(4,4+x))$$
Here $P(X\geq4)=e^{-4}$ and for RHS we discern:
$x<0$ then $P(X\geq\max(4,4+x))=P(X\geq4)>0$ and leading to $$P(X-4\geq x\mid Y\geq5)=1$$
$x\geq0$ then $P(X\geq\max(4,4+x))=P(X\geq4+x)=e^{-4+x}$ and leading to:$$P(X-4\geq x\mid Y\geq5)=e^{-x}$$
Actually this can be found more directly on base of memorylessness of exponential distribution stating that $P(X>s+t\mid X>s)=P(X>t)$.