Question 1172514
Let's solve this problem step-by-step.

**Given:**

* Joint distribution function:
    * F(x, y) = (1 - e^(-x))(1 - e^(-y)) for x > 0, y > 0
    * F(x, y) = 0 elsewhere

**1. Compute P({X > 1} ∪ {Y > 1})**

We'll use the formula:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

In our case:

* A = {X > 1}
* B = {Y > 1}

* **P(X > 1):**
    * P(X > 1) = 1 - P(X ≤ 1) = 1 - lim(y→∞) F(1, y)
    * P(X > 1) = 1 - (1 - e^(-1)) = e^(-1)

* **P(Y > 1):**
    * P(Y > 1) = 1 - P(Y ≤ 1) = 1 - lim(x→∞) F(x, 1)
    * P(Y > 1) = 1 - (1 - e^(-1)) = e^(-1)

* **P(X > 1, Y > 1) = P(X > 1 ∩ Y > 1):**
    * P(X > 1, Y > 1) = 1 - P(X ≤ 1) - P(Y ≤ 1) + P(X ≤ 1, Y ≤ 1)
    * P(X > 1, Y > 1) = lim(x→∞, y→∞) F(x,y) - lim(x→∞,y=1)F(x,y) - lim(x=1,y→∞) F(x,y) + F(1,1)
    * P(X > 1, Y > 1) = 1 - (1-e^-1) - (1-e^-1) + (1-e^-1)(1-e^-1)
    * P(X > 1, Y > 1) = 1 - 1 + e^(-1) - 1 + e^(-1) + 1 - 2e^(-1) + e^(-2)
    * P(X > 1, Y > 1) = e^(-2)

* **P(X > 1 ∪ Y > 1):**
    * P(X > 1 ∪ Y > 1) = P(X > 1) + P(Y > 1) - P(X > 1, Y > 1)
    * P(X > 1 ∪ Y > 1) = e^(-1) + e^(-1) - e^(-2)
    * P(X > 1 ∪ Y > 1) = 2e^(-1) - e^(-2)

**2. Compute P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2)**

* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = F(2, 2) - F(1, 2) - F(2, 1) + F(1, 1)
* F(2, 2) = (1 - e^(-2))(1 - e^(-2))
* F(1, 2) = (1 - e^(-1))(1 - e^(-2))
* F(2, 1) = (1 - e^(-2))(1 - e^(-1))
* F(1, 1) = (1 - e^(-1))(1 - e^(-1))

* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = (1 - e^(-2))(1 - e^(-2)) - (1 - e^(-1))(1 - e^(-2)) - (1 - e^(-2))(1 - e^(-1)) + (1 - e^(-1))(1 - e^(-1))
* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = (1 - e^(-2) - (1-e^-1)) * (1-e^-2 - (1-e^-1))
* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = (e^(-1) - e^(-2)) * (e^(-1) - e^(-2))
* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = (e^(-1) - e^(-2))^2

**Summary**

* P({X > 1} ∪ {Y > 1}) = 2e^(-1) - e^(-2)
* P(1 ≤ X ≤ 2, 1 ≤ Y ≤ 2) = (e^(-1) - e^(-2))^2