Here's my work for part a. I could use clarification on part b and d. Is part d the same as part a ($E[A_n] = E[Y]$) ?
a) $$E[Y_n] = E[\frac{X_n}{2^n}]$$ ($X$'s are iid so...) $$= \frac{E[X]}{2^n} = \frac2{2^n}$$
$$Var(Y_n) = \frac{Var(X_n)}{4^n} = \frac9{4^n}$$
b) $E[T_n] = E[\sum_{i=0}^n Y_i] = E[1 + \frac12 + \frac14 + ... + \frac2{2^n}] = E[\frac1{1-\frac12}] = 2$
Is this correct?
What you have for b is wrong this is because $$E[T_{n}]=E\left[\sum_{i=1}^{n}Y_{i}\right]=\sum_{i=1}^{n}E[Y_{i}]=\sum_{i=1}^{n}\frac{2}{2^{i}}=\sum_{i=1}^{n}\frac{1}{2^{i-1}}=\sum_{i=0}^{n-1}\frac{1}{2^{i}}$$
Now here you also attempted to use fact about a geometric series, but remember n is finite here (it is not approaching infinity like it does in a geometric series). Thus we actually have (which is derived here (http://en.wikipedia.org/wiki/Geometric_series#Proof_of_convergence) $$=\frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}}=2\left(1-\frac{1}{2^{n}}\right)$$
For further use of this, i guess you would call it, "finite geometric sum" the property is that $$\sum_{i=0}^{n-1}ar^{i}=a\frac{1-r^{n}}{1-r}$$ where you must make sure indices and such are exactly the same form to use this exact result.