An integral in Peskin's Quantum Field Theory P. 27
In those paragraphs, P&S want to obtain the behaviour of the amplitude $D(x-y)$ in the limit $r \to \infty$. Starting from the integral $$ \tag{1} \label{int} \frac{-i}{2(2\pi)^2 r} \int_{-\infty}^\infty \text{d}p \frac{p e^{ipr}}{\sqrt{p^2 +m^2}}, $$ you can see that a limit for $r \to \infty$ is not well defined, since the exponential has an oscillating behaviour. Also, the final form of the integral can be recognized as a modified Bessel function of the second kind. See this Wikipedia page for the asymptotic expansion of this function. The modification of the integration contour is a way to obtain a more straightforward integral. I hope this answers your question 1.
About point 2., in order to evaluate this integral, P&S apply the Cauchy's theorem defining the integrand as a function of the complex variable $p$. The square root in the denominator causes the branch cut in Figure 2.3. When rewriting the contour integral as $$ \tag{2} \label{1} \int_{i \infty}^{i m} \dots + \int_{i m}^{i \infty} \dots $$ we have to be careful in writing the integrands, since the branch cut generates a discontinuity between the two sides of the cut. In fact the square root in the complex plane is a multi-valued function, and the argument of $p^2+m^2$ shifts of $2\pi$ when passing from the right of the branch cut to the left. Thus $$ \sqrt{p^2 + m^2} = \cases{|p^2+m^2|^{1/2} \exp\left(i\frac{\arg(p^2+m^2)}{2}\right), \\ |p^2+m^2|^{1/2} \exp\left(i\frac{\arg(p^2+m^2)+ 2\pi}{2}\right) = - |p^2+m^2|^{1/2} \exp\left(i\frac{\arg(p^2+m^2)}{2}\right),} $$ where the first value is at the right and the second at the left of the branch cut. You see that the difference is a minus sign.
Therefore, making the substitution $p = i \rho$, \eqref{1} becomes $$ i\int_\infty^m \text{d} \rho \frac{\rho e^{-r \rho}}{- \sqrt{\rho^2 - m^2}} + i\int_m^\infty \text{d} \rho \frac{\rho e^{-r \rho}}{\sqrt{\rho^2 - m^2}}, $$ which finally gives $$ 2i \int_m^\infty \text{d} \rho \frac{\rho e^{-r \rho}}{\sqrt{\rho^2 - m^2}}. $$ Substituting this in \eqref{int} you find your result. Note that if you had chosen a different contour, for example the analogous of the "pushed" contour of Figure 2.3 but with the U-turn at $p = 0 + i 0$, you would not have the phase difference in the integrands on the lines $(im, 0)$ and $(0, im)$, so these two pieces would have cancelled each other (see \eqref{1}) and you would have obtained the same result with $m$ as lower integration limit.