1. We find $\Pr(Y\le y)$. This is $\Pr(a+(b-a)X)\le y$, which is $\Pr\left(X\le \frac{y-a}{b-a}\right)$.
If $y-a\le 0$, that is, if $y\le a$, this probability is $0$.
If $\dfrac{y-a}{b-a}\ge 1$, that is, if $y\ge b$, this probability is $1$.
And finally, the interesting part. If $a\lt y\lt b$, this probability is $\dfrac{y-a}{b-a}$.
The three sentences above describe the cumulative distribution function of $Y$. For the density, differentiate. We get density $\dfrac{1}{b-a}$ on $(a,b)$ and $0$ elsewhere. Note that $Y$ has uniform distribution on $(a,b)$ (or equivalently, $[a,b]$).
2. a) Our distributions are continuous. So, as the problem is currently stated, the probability is $0$.
For b), each of $X_1$ and $X_2$ has density function of the shape $e^{-t/1000}$ when $t\gt 0$. So the probability the first is alive after $1200$ hours, by integration, is $e^{-1200/1000}$. The same is true for the second.
Because $X_1$ and $X_2$ are independent, the probability both components are still alive after $1200$ hours is the product of the individual probabilities, that is, $e^{-2400/1000}$. If we interpret "device fails after $1200$ hours" as meaning that the lifetime of the device is at least $1200$, that gives the answer to the question.
The likelihood function is $$L(\theta|\mathbb x)=\begin{cases}\dfrac{1}{\theta ^n},\,\,\,\theta \le x_i \le 2\theta ,\forall i\\0,\,\,\,\,\,\,\,\,\text{otherwise}\end{cases}$$ $$=\begin{cases}\dfrac{1}{\theta ^n},\,\,\,\theta \le x_{(1)} \le x_{(n)} \le2\theta \\0,\,\,\,\,\,\,\,\,\text{otherwise}\end{cases}$$
For $\theta \ge \dfrac {x_{(n)}}{2},L(\theta|\mathbb x)=\dfrac{1}{\theta ^n}$ is a decreasing function in $\theta$. Hence MLE of $\theta$ is $\color{blue}{\hat\theta=\dfrac{X_{(n)}}{2}}$.
Best Answer
We will need the joint comulatice distribution function $F_{Y_1,Y_2}(u,v)=P(Y_1<u,Y_2<v).$
Now,
$$P(Y_1<u,Y_2<v)=P(Y_1<u,Y_2<v, X_1\le X_2)+P(Y_1<u,Y_2<v ,X_2<X_1)=$$ $$=P(X_1<u,X_2<v, X_1\le X_2)+P(X_2<u,X_1<v ,X_2<X_1).$$
First, $$P(X_1<u,X_2<v, X_1\le X_2)= \begin{cases} -\frac{u^2}{2}+uv,&\text{ if }& 0\le u\le v \le 1\\ \frac{v^2}{2},&\text{ if }& 0\le v<u \le1 \end{cases}.$$
The figure below explains how I calculated the two pieces of surface area:
Second,
$$P(X_2<u,X_1<v ,X_2<X_1)=P(X_2<u,X_1<v, X_2\le X_1)= \begin{cases} -\frac{u^2}{2}+uv ,&\text{ if }& 0\le u\le v \le 1\\ \frac{v^2}{2},&\text{ if }& 0\le v<u \le1 \end{cases}.$$ (A figure similar to the one above would explain the latter result.) We can calculate the joint cdf: $$F_{Y_1,Y_2}(u,v)=$$$$=P(X_1<u,X_2<v, X_1\le X_2)+P(X_2<u,X_1<v ,X_2<X_1)=$$
$$=\begin{cases} -u^2+2uv,&\text{ if }& 0\le u\le v \le 1\\ v^2,&\text{ if }& 0\le v<u \le1 \end{cases}.$$
The joint pdf can be calculated the following way: $$f_{Y_1,Y_2}(u,v)=\frac{{\partial} ^2}{\partial u \partial v}F_{Y_1,Y_2}(u,v)=\begin{cases} 2,&\text{ if }& 0\le u\le v \le 1\\ 0,&\text{ otherwise. } \end{cases}$$
Then we will need the following marginal density:
$$f_{Y_2}(v)=\int_0^vf_{Y_1,Y_2}(u,v)du=2\int_0^v\ du=2v;\ 0\le v\le 1.$$
To answer the first question: By definition
$$f_{Y_1 \mid Y_2=v}(u)=\frac{f_{Y_1,Y_2}(u,v)}{f_{Y_2}(v)}=\begin{cases} \frac{1}{v},&\text{ if }& 0\le u\le v \le 1\\ 0,&\text{ else. } \end{cases} \tag 1$$
To answer the second question: First let's calculate the cdf of $Y_2-Y_1$. By definition
$$F_{Y_2-Y_1}(x)=P(Y_2-Y_1<x)=E[P(Y_2-Y_1<x \mid Y_2)]. $$
Then
$$P(Y_2-Y_1<x \mid Y_2=v)= \begin{cases} 0,&\text{ if } x<0\\ P(Y_1>v-x\mid Y_2=v),& \text{ if } 0\le x \le v \le 1\\ 1,&\text{ if } x>v \end{cases}$$
where $0\le v \le 1$.
Considering $(1)$ we have for $0\le x \le v \le 1$
$$P(Y_1>v-x\mid Y_2=v)=\frac 1v\int_{v-x}^v\ du=\frac xv.$$
So, for a given $0\le x \le 1$
$$P(Y_2-Y_1<x \mid Y_2=v)=\begin{cases} 1,& \text{ if } 0\le v\le x\\ \frac xv,&\text{ if } x\le v\le 1.\\ \end{cases}$$
As a result, for $0\le x \le 1$
$$F_{Y_2-Y_1}(x)=\int_0^1P(Y_2-Y_1<x \mid Y_2=v)f_{Y_2}(v)\ dv=$$ $$=2\int_0^x v \ dv + 2x\int_x^1 \ dv=2x-x^2.$$
Finally, $$f_{Y_2-Y_1}(x)=\begin{cases} 2-2x,& \text{ if } 0\le x \le 1 \\ 0,&\text{ otherwise. } \end{cases}$$