How do you calculate the probability of the difference between two normal distribution If I have two normally distributed graphs, one with a mean of 70 and standard deviation of 3, and the other with a mean of 74 and standard deviation of 4.5, how would I go about calculating the probability of a random point from the first distribution being 2 larger than a random point on the second distribution?

I will show how to work a similar problem, and let you make the required
changes for your own problem. I hope you can use my work for the alternative
problem as an outline for doing yours. Please check each step with formulas
and theorems in your text or notes.

Note: @DavidQuinn is correct, that the wording of
  your problem is faulty (unless it is intended to be trivial); to make sense it needs to say
  something like "larger by at least two." Also, you need to say that
  $X$ and $Y$ are independent random variables. Please look into that.

Alternative Problem: Suppose $X \sim \mathsf{Norm}(70, 3)$ and (independently)  $Y \sim \mathsf{Norm}(74, 4).$ [The second argument in my notation $\mathsf{Norm}$ is the standard deviation.]
Then find the distribution of $D = X - Y$ and $P(D > 2).$
Mean of the sum (or difference) of two random variables. A general formula states that $E(aX + bY) = aE(X) + bE(Y).$ In the
current problem $a = 1$ and $b = -1,$ so $E(X+Y) = 70-74 = -4.$
Variance of the sum (or difference) of two independent random variables. Another general formula states that for independent random variables
$X$ and $Y,$ one has $Var(aX + bY) = a^2Var(X) + b^2Var(Y).$ In the
current problem, this gives $Var(X - Y) = 3^2 + 4^2 = 25,$ so
$SD(X - Y) = \sqrt{25} = 5.$ Notice that the variances are added
even though one random variable is subtracted from the other.
Distribution of the sum of independent normal random variables. A third general result states that the sum (or difference) of two
independent normal random variables is is another normal random
variable with mean and SD given by the two relationships above.
Thus $D = X - Y \sim \mathsf{Norm}(-4, 5).$  
The last step is to use software or printed standard normal tables to find $P(D > 2).$ [If you use standard normal tables, you will have to standardize $D.$
I assume you have done that before.] The answer to the current problem (from software) is $P(D > 2) = 0.1149.$ Because $Y$ has a larger mean than $X,$ it
is not surprise that this probability is considerably smaller than $1/2.$
