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.
Best Answer
I suppose that the exponential distribution is independent of the end of the experiment. Let $X$ be the end of the experiment and $Y$ be the time at which the device turns of. You are looking at the probability of the event $X\leq Y$, that is \begin{align*} \mathbb P[X\leq Y] &= \int_2^6 \int_x^\infty f_{X,Y}(x,y) dy dx\\ &= \int_2^6 \int_x^\infty \frac{1}{4}\cdot \frac{1}{4}e^{-\frac{1}{4}y} dy dx\\ &= \frac{1}{4}\int_2^6 e^{-\frac{1}{4} x} dx\\ &= e^{-\frac{1}{2}} - e^{-\frac{3}{2}} \end{align*}