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.
Since $y_1 < y_2$
$P(Y_1 \leq y_1 , Y_2 \leq y_2) = P(\min(X_1, X_2) \leq y_1 \text{ and } \max(X_1, X_2) \leq y_2) = P( \{ X_1 \leq y_1 \text{ or } X_2 \leq y_1 \} \text{ and } X_1 \leq y_2 \text{ and } X_2 \leq y_2) = P(\{X_1 \leq y_1 \text{ and } X_2 \leq y_2 \} \text{ or } \{X_1 \leq y_2 \text{ and } X_2 \leq y_1 \})$
Now, just use the fact that $P(A \text{ or } B) = P(A) + P(B) - P(A\text{ and } B)$.
Best Answer
We look the easier case first, $P(a_1 \leq Y_1 \leq a_2,b_1\leq Y_2 \leq b_2) = \int_{b_1}^{b_2} \int_{a_1}^{a_2}f(y_1,y_2)dy_1dy_2$. You need to compute the integral over the rectangle region $[a_1,a_2]\times[b_1,b_2]$.
Back to your question, $P(2T_1+T_2 < x)$, and $t_1,t_2>0$, so what you need to do is to find the integral region on $(t_1,t_2)$-plane. The region of $t_1$ goes from the line $t_1=0$ to $t_1=\frac{1}{2}(t_2-x)$, and when $t_1$ is given, the region of $t_2$ is from $0$ to $x$.