MGF, Calculate the standard deviation of $X + Y$ . Method. 
Two claimants place calls simultaneously to an insurers claims call
  center. The times X and Y , in minutes, that elapse before the
  respective claimants get to speak with call center representatives are
  independently and identically distributed. The moment generating
  function for both $X$ and $Y$ is
  $$ M(t) = \left( \frac{1}{1-1.5t}\right)^2, \quad t< \frac{2}{3}.$$
  Calculate the standard deviation of $X + Y$ .

I would like to know what approaches I need to use to solve this exercise. I am practicing for an actuarial exam and I am looking for the method that will minimize the amount of time spent on these types of problems.
Is it correct this answer, (I used the coment below my post):
X and Y are Erlang(2, 2/3) and independent, then var(x+y)=var(x)+var(y)+2 cov(x,y)
because are independent,
Var(x+y)= 2 var(x)= 2 n * (alpha)^2  =2 (2) (2/3)^2

Then,
satandar deviation= 2 * (alpha) = 4/3

 A: I would first write $$M(t) = (1-3t/2)^{-2}$$ so that $$M'(t) = (-3/2)(-2)(1-3t/2)^{-3} = 3(1-3t/2)^{-3}.$$  Thus $M'(0) = 3$.  Next, $$M''(t) = (-3/2)(-3)(3)(1-3t/2)^{-4} = \frac{27}{2}(1-3t/2)^{-4},$$ so $M''(0) = \frac{27}{2}$.  This immediately gives us, for example, $$\operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2 = \frac{27}{2} - 3^2 = \frac{9}{2},$$ and since $X$ and $Y$ are independent and identically distributed, the variance of their sum is equal to the sum of their variances.  Hence $$\sqrt{\operatorname{Var}[X+Y]} = \sqrt{\frac{9}{2} + \frac{9}{2}} = 3.$$

Why did I do it this way, when advice was given to recognize the MGF as belonging to an Erlang (Gamma) distribution?  Because while time is of the essence for the actuarial examinations, accuracy is more important.  Do not forget this.  It is better to take a little extra time and be sure you have answered the question correctly, rather than to rely on shortcuts and memorized facts that do not so readily admit alternative means of verification.
As you can see, the answer you got from the "shortcut" method is incorrect: this is because there are two commonly used parametrizations of the gamma distribution; one by shape $\alpha$ and scale $\theta$:  $$f_X(x) = \frac{x^{\alpha-1} e^{-x/\theta}}{\theta^\alpha \Gamma(\alpha)}, \quad x > 0,$$ and one by shape $a$ and rate $b$:  $$f_X(x) = \frac{b^a x^{a-1} e^{-bx}}{\Gamma(a)}, \quad x > 0.$$  They are related through the relationships $$a = \alpha, \quad b\theta = 1.$$  But if you simply memorize, you can very easily mix up the two:  under the first parametrization, $$\operatorname{Var}[X] = \alpha \theta^2,$$ as you used.  Under the second parametrization, $$\operatorname{Var}[X] = a/b^2.$$  So when the comment was given to recognize the MGF as belonging to a gamma distribution, which variance formula did you use?  You ended up picking the wrong one, because when one person writes $$\operatorname{Gamma}(2,2/3),$$ you don't know whether that is a shape/rate parametrization, or a shape/scale parametrization.  You'd have to know how the parametrization choice also relates to the resulting MGF:  for shape/scale, the MGF is $$M_X(t) = (1-\theta t)^{-\alpha}.$$  For shape/rate, the MGF is $$M_X(t) = (1-t/b)^{-a}.$$  See how having to remember which is which can be more confusing than simply using what you are given to calculate the answer?  This is the sort of trap that the designers of the actuarial examinations love to set, because the purpose of the test is not to see whether examinees can memorize the formulas.  The purpose of the test is to gauge true understanding.
The key to succeeding on the preliminary exams is to learn how to be fast in your calculations, and then to take extra time to find ways to check that your calculation is correct, usually by doing the problem in an alternative way, or draw upon other facts to corroborate that you recalled the necessary concepts correctly.
A: An even quicker way than what heropup has would be to use the fact that for a random variable $W$ with MGF $M_{W}$, $$\dfrac{\text{d}^2}{\text{d}t^2}\ln(M_{W}(t))\Large|_{\normalsize t=0} \normalsize= \text{Var}[W]\text{.}$$
We know that $M_{X}(t) = M_{Y}(t) = \left(\dfrac{1}{1-1.5t}\right)^2\text{, } t < 2/3$.
Since $X$ and $Y$ are independent, it follows that $$M_{X+Y}(t) = M_{X}(t)M_{Y}(t) = \left(\dfrac{1}{1-1.5t}\right)^4\text{.}$$
Furthermore, it is easily seen that $$\ln M_{X+Y}(t) = -4\ln(1-1.5t)\text{.}$$
The first derivative is $$-4\left(\dfrac{1}{1-1.5t}\right)(-1.5)$$
The second derivative is $$-4\left(\dfrac{1}{1-1.5t}\right)^2(-1.5)^2(-1) $$
and evaluated at $t = 0$, we have $\text{Var}[X+Y] = -4(-1.5)^2(-1) = 9$. Thus, the standard deviation is $\sqrt{9} = 3$.
A: We have the following:
Theorem Let $X$ be an $\mathbb{R}$-valued random variable for which the $M_X(t)=\mathbb{E}(e^{tX})$ exists and is finite in a neighbourhood of $0$, then:
$$
\mathbb{E}X^n=M^{(n)}_X(0)
$$
From this, it easily follows that
$$
\text{Var}X=\mathbb{E}X^2-(\mathbb{E}X)^2=M_X''(0)-(M'_X(0))^2
$$
Theorem If $X$ and $Y$ are independent random variables and their respective mgf exists, then
$$
M_{X+Y}(t)=M_X(t)M_Y(t)
$$
