If you have learned how to visualize the concept of conditional densities, the answer to part (c) (and indeed parts (a) and (b) as well) can be obtained with a little bit visualization and hardly any calculation. I will explain below, but it takes longer to write down than to simply draw a simple sketch and deduce the answers. A picture is indeed worth the next thousand words!
The joint density can be thought of as a right prism of height $1$ with triangular base sitting on the $x$-$y$ plane. The base has vertices are $(0,0)$, $(1,0)$ and $(1,2)$, and thus area $1$. The conditional density of $Y$ given that $X$ has taken on value $x$ is proportional to the cross-section of this prism by the plane at $x$. For $0 < x < 1$, this cross-section is a rectangle with base of length $2x$, and so the conditional density of $Y$ is a uniform density $U(0,2x)$, that is, $f_{Y\mid X}(y\mid X=x) = \begin{cases}\frac{1}{2x}, & 0 \leq y \leq 2x,\\ & \\ 0, & \text{otherwise.}\end{cases}$ Since the conditional density is uniform on $(0,2x)$, it follows immediately that $E[Y \mid X=x] = x$.
Turning to the matter of calculating $\rho$, note that by very similar visualization, the conditional density of $X$ given $Y = y$ is a uniform density on $(y/2, 1)$ and thus $E[X \mid Y = y] = 1/2 + y/4$, the midpoint of the base of the uniform density.
Now, $E[Y\mid X = x]$ and $E[X\mid Y = y]$ are the minimum-mean-square-error (MMSE) estimators for $Y$ and $X$ respectively given the value of the other random variable, and since these are linear functions, they are also the MMSE linear estimators for $Y$ and $X$ respectively. Now, the MMSE linear estimators are lines through the mean point $(\mu_X,\mu_Y)$ given by $\begin{align*} \frac{y-\mu_Y}{\sigma_Y} &= \rho\frac{x-\mu_X}{\sigma_X}\qquad \equiv \quad y = x\\ \frac{x-\mu_X}{\sigma_X} &= \rho \frac{y-\mu_Y}{\sigma_Y} \qquad \equiv \quad x = \frac{1}{2}+\frac{y}{4} \end{align*}$ Clearing fractions, we see that $\rho\frac{\sigma_Y}{\sigma_X} = 1$ and $\rho\frac{\sigma_X}{\sigma_Y} = \frac{1}{4}$, giving $\rho^2 = \frac{1}{4}$ and $\rho=\frac{1}{2}$. (The solution $\rho = -\frac{1}{2}$ can be discarded since we know that $\rho$ is positive.)