Cov and joint pdf question (struggling)

I have these 2 questions that I struggled to do ... I wanted to ask my fellow yahooers ... Hopefully, someone can help me out. I posted the picture of the both questions so I appreciate the help.

Thank you,

http://www.picsend.net/images/313653probabilit.jpg

E(Y)
= E(X1 + 2X2 + 3X3)
= E(X1) + 2E(X2) + 3E(X3)
= (μ) + 2(μ) + 3(μ)
= 6μ

V(Y)
= V(X1 + 2X2 + 3X3)
= 1² V(X1) + 2² V(X2) + 3² V(X3) + 2(1)(2) Cov(X1, X2) + 2(1)(3) Cov(X1, X3) + 2(2)(3) Cov(X2, X3)
= V(X1) + 4 V(X2) + 9 V(X3) + 4 Cov(X1, X2) + 6 Cov(X1, X3) + 12 Cov(X2, X3)
= (σ²) + 4 (σ²) + 9 (σ²) + 4 (1) + 6 (1) + 12 (1)
= 14σ² + 22

=================

Show the marginal pdf of X1, X2, X3
X1 ~g1(X) = ∫(0 to ∞) ∫(0 to ∞) f(x1, x2, x3) dx2 dx3 = 3e^(-3x1)
X2 ~g2(X) = ∫(0 to ∞) ∫(0 to ∞) f(x1, x2, x3) dx1 dx3 = 3e^(-3x2)
X3 ~g3(X) = ∫(0 to ∞) ∫(0 to ∞) f(x1, x2, x3) dx1 dx2 = 3e^(-3x3)

Since g1(x) * g2(x) * g3(x) = f(x1, x2, x3)
They are independent

And they are in the form of an exponential function with ß = 3
Thus their mgf is (1 - 3t)^-1

And
mgf(X1 + X2)
= (1 - 3t)^-1 * (1 - 3t)^-1
= (1 - 3t)^-2
Which in the form of mgf of a gamma distribution with alpha = 2, beta = 3

Similarly
mgf(X1 + X2 + X3)

= (1 - 3t)^-1 * (1 - 3t)^-1 * (1 - 3t)^-1
= (1 - 3t)^-3
Which in the form of mgf of a gamma distribution with alpha = 3, beta = 3