Hello,

By definition, , where denotes the n-th derivative of the mgf.

Specifically, and

For Var(X), recall the formula I put before : E(X²)-[E(X)]²

For question C., I don't know... Your mgf is the mgf of an exponential distribution with parameter 1/3, which would finish the question, by using the cumulative distribution function.

I'm looking for a way without "recognizing" the mgf, but it looks like there isn't any.