I will point out that contrary to the question title, not two, but three pieces of information have been communicated in the question: (1) the fact that the scores are normally distributed, (2) the mean of the scores, and (3) the standard deviation of the scores. By far the most important is that first piece of information, so don't overlook that. (Indeed, if only the mean and standard deviation were given, then the question could not be answered.)
For practical purposes, one turns to statistical tables or technology which give the area under selected parts of a normal curve, which is the same as the percent/probability of elements in that interval (i.e., the "integral" from calculus). It is convenient to first standardize the distribution in question, transforming it to the normal curve with mean 0 and standard deviation 1 (which has the same relative areas, and thus requires only a single table to look-up, or simplified programming within the computer application).
So for a percent-less-than question, one would standardize and then look up in a normal curve left-area table (or else use technology). For the question here, the standard score for the cut is $z = \frac{x - \mu}{\sigma} = \frac{44 - 60}{8} = -2$. Looking up this z-score in a normal table like here gives the area of 0.0228, that is 2.28%.
This is a very fundamental technique when dealing with common probability distributions (esp. the normal curve), and if you're in a class on the subject it should be well-laid out in a chapter on this subject. For this particular problem with $z = -2$, some will even be able to answer it mentally by remembering the Empirical Rule (or making a sketch); knowing that 95.44% of the area is within $z = \pm 2$, the remainder in the outside tails is 4.56%, with exactly half in the left-hand tail, that is, $4.56\% / 2 = 2.28\%$.
Let random variable $X$ have normal distribution with mean $\mu$ and standard deviation $\sigma$. Then
$$\Pr(X\gt a)=\Pr(X-\mu\gt a-\mu)=\Pr\left(\frac{X-\mu}{\sigma}\gt \frac{a-\mu}{\sigma}\right)=\Pr\left(Z\gt \frac{a-\mu}{\sigma}\right),$$
where $Z$ is standard normal.
In our case, we have $\mu=100$, and $\sigma=12$, and $a=104$. So we want
$$\Pr\left(Z\gt \frac{4}{12}\right).\tag{1}$$
Remark: To evaluate (1) numerically, one can use software, or, if one is old-fashioned, one uses tables of the standard normal distribution. In fact we do not really need to work with the standard normal $Z$, for there are programs, and online normal distribution calculators, that will evaluate $\Pr(X\gt a)$ directly if you input $\mu$, $\sigma$, and $a$.
Best Answer
You have vehicle speeds $X \sim \mathsf{Norm}(\mu = 36, \sigma = 2)$ and you seek $$P(X > 40) = P\left(\frac{X - \mu}{\sigma} > \frac{40-36}{2}\right) = P(Z > 2) = 0.02275,$$ where $Z$ has the standard normal distribution and the probability can be found using printed normal tables or software.
You do not define what you mean by $R,$ but the numerical answer is OK.
Note: Using some kinds of statistical software you can skip the 'standardization step' and get the answer directly. In R statistical software, for example, you could use
In Minitab 16 you can get $P(X \le 40)$ and then subtract from $1.$