Let X and Y be independent exponential random variables with shared parameter $\lambda$. Define the random variables U = Y and V = X+ XY

I'm asked to find E(E(V|U)), Var(E(V|U)), and E(Var(V|U)). I think the answer to the first one is $\lambda + \lambda^2$. Here I used E(V)=E(E(V|U)). I'm not sure about the other two though. I tried to do the transformation involving the Jacobian to find the joint pdf of V and U, but I can't find an explicit expression for the marginal distributions, so I can't find an explicit expression for the conditional distribution of V|U. Is there another way to do this?

No transformations are necessary. If $U = Y$ and $V = X + XY$, then $V = (1+U)X$ and we simply have

$$\operatorname{E}[V \mid U] = \operatorname{E}[(1+U)X \mid U] = (1+U)\operatorname{E}[X] = (1+U)\lambda.$$

Similarly,

$$\operatorname{Var}[V \mid U] = (1+U)^2 \operatorname{Var}[X] = (1+U)^2 \lambda^2.$$

Now all that remains is to take the variance and expectations with respect to $U$:

$$\operatorname{E}[\operatorname{E}[V \mid U]] = \lambda \operatorname{E}[1 + U] = \lambda (1 + \lambda),$$

and

$$\operatorname{Var}[\operatorname{E}[V \mid U]] = \lambda^2 \operatorname{Var}[1+U] = \lambda^2 \lambda^2 = \lambda^4,$$

and

$$\operatorname{E}[\operatorname{Var}[V \mid U]] = \lambda^2 \operatorname{E}[(1 + U)^2] = \lambda^2 (1 + 2 \lambda + \operatorname{Var}[U] + \operatorname{E}[U]^2) = \lambda^2 (1 + 2\lambda + 2\lambda^2).$$