The pdf of $Y$ is obtained by taking the joint pdf of $(X,Y)$ and marginalizing $X$ out. That is:
$$f_Y(y)=\int_{-\infty}^\infty f_{X,Y}(x,y) dx.$$
The joint pdf of $(X,Y)$ is the product of the conditional pdf $f_{Y|X}(y|x)$ and the pdf of $X$, $f_X$. (If this seems weird to you, it is basically analogous to the familiar identity $P(A \cap B)=P(A \mid B) P(B)$.) That is:
$$f_{X,Y}(x,y)=f_{Y|X}(y|x) f_X(x).$$
You have these two pdfs, so with this and some calculus you can do part 1. Once you have the joint pdf you can compute the covariance with some more calculus, so you can do part 2.
The one thing you seem to be having trouble reading is the conditional pdf. The problem is trying to tell you that $f_{Y|X}(y|x)$, for each fixed $x$, is the pdf of a normal r.v. with mean $x$ and variance $1$.
One way to understand the calculation is to recall that for a gamma distribution with shape $\alpha$ and scale $\beta$, $$f_X(x) = \frac{x^{\alpha-1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)}, \quad x > 0.$$ The denominator, being independent of $x$, suggests that $1/(\beta^\alpha \Gamma(\alpha))$ is the required multiplicative factor for the density such that $$\int_{x=0}^\infty f_X(x) \, dx = 1.$$ In other words, $$\beta^\alpha \Gamma(\alpha) = \int_{x=0}^\infty x^{\alpha - 1} e^{-x/\beta} \, dx.$$ This holds true for any $\alpha, \beta > 0$. Now, with this in mind, $$\operatorname{E}[X^\nu] = \int_{x=0}^\infty x^\nu f_X(x) \, dx = \int_{x=0}^\infty \frac{x^{\nu + \alpha - 1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)} \, dx = \frac{\beta^{\nu + \alpha} \Gamma(\nu + \alpha)}{\beta^\alpha \Gamma(\alpha)}\int_{x=0}^\infty \frac{x^{\nu + \alpha - 1} e^{-x/\beta}}{\beta^{\nu + \alpha}\Gamma(\nu + \alpha)} \, dx.$$ This is what the provided solution does, but the motivation for doing so should now be clear, because now the integrand represents a gamma density with shape parameter $\nu + \alpha$, and rate $\beta$. Therefore, its integral over its support is also $1$, provided $\nu + \alpha > 0$. It follows that $$\operatorname{E}[X^\nu] = \frac{\beta^\nu \Gamma(\nu + \alpha)}{\Gamma(\alpha)}, \quad \nu > -\alpha,$$ as claimed.
Best Answer
$\alpha$ is the lower end of the uniform distribution while $\beta$ is the upper end. All possible values between them are equally likely
Hint:
$$E[X]=\frac{\alpha+\beta}{2}$$
since the mean is also the mid point, so you want to calculate
$$E\left[\left(X-E[X]\right)^r\right]=\dfrac1{\beta-\alpha}\int_\alpha^\beta \left(x-\frac{\alpha+\beta}{2}\right)^r \, dx$$