I have an exercise stating what is written in the title-box.
"Cx on [0,3]"
I am not sure about what to do.
The function in your lower image is the Gaussian probability density function.
That page will tell you everything you need to know to answer your problem and more.
You need to learn this distribution as it is arguably the most important one there is.
I see no reason to think that the given problem has anything to do with the Gaussian distribution you give below. The problem asks you to first find the value of "C" so that f(x)= Cx, $\displaystyle 0\le x\le 3$ is a probability distribution. The key point is the total probability, $\displaystyle \int_0^3 Cx dx$ is equal to 1. So, do the integration, set it equal to 1, and solve for C.
For part (b), you are asked to find the mean, $\displaystyle \mu$, and standard deviation, $\displaystyle \sigma$. The mean is the expected value of x itself, so is $\displaystyle \mu= \int_0^3 x(Cx)dx= \int_0^3 Cx^2 dx$. The deviation is $\displaystyle \sigma= \int_0^3 (x- \mu)^2Cx dx= \int_0^3 Cx^3 dx+ 2\int_0^3 \mu C x^2 dx+ \int_0^3 \mu^2C dx$.
Once you have found those, part (c) is $\displaystyle P(\mu-\sigma\le X\le \mu+ \sigma)= n\int_{\mu-\sigma}^{\mu+ \sigma} Cx dx$. (What you have written says that this is the probability the random variable, X, is greater than one standard deviation from the mean but, in fact, it is the probability it is within one standard deviation.)
But I am puzzled by your whole post. This is a fairly complicated problem, not requiring any difficult computation but certainly requiring that you understand several basic concepts. Yet, you seem to be saying that you do not know what any of the words in the problem mean! If, as you seem to be saying, you have never seen the idea that the total probability for a probability distribution must be 1, or what a "mean" or "standard distribution" are, I can't imagine why you were given a problem like this.