The first question seems more complex to me than the second question. An answer to the second question arises naturally out of the work needed to answer the first question.
Imagine the galaxy as the sum of an onion-like collection of infinitesimally thin spherical "shells" of volume, each with a thickness $dr$, surface area $A = 4 \pi r^2$, and thus volume $dV = A \, dr = 4 \pi r^2 \,dr$.
As $m = \rho V$, it follows that the infinitesimally small mass of each shell, $dm$, is $dm = \rho(r) \,dV$. Since the volume of the shall varies with its distance from the centre ($dV = 4\pi r^2 \,dr$), we rephrase $dV$ in terms of $r$ with the intention of integrating both sides - this is especially because our expression for $\rho$ varies with $r$, too.
As such, we find an expression for the mass of the galaxy at a radius $r$, $m(R)$:
$$m(R) = \int_{0}^{R} dm = \int_{0}^{R} \rho(r) dV = \int_{0}^{R} \rho(r) 4 \pi r^2\, dr$$
Where we want to evaluate the integral on the right-hand side of the above equation.
$$4 \pi \rho_c \int^{R}_{0} e^{-\frac{r}{r_0}} r^2\, dr$$
This can be done through integration by parts; take $r^2$ as the expression to differentiate, and $e^{-\frac{r}{r_0}}$ as the expression to integrate. We find that
$$4 \pi \rho_c \int^{R}_{0} e^{-\frac{r}{r_0}} r^2\, dr = 4\pi\rho_c \left[ -e^{-\frac{r}{r_0}} (r^2r_0 + 2r r_0^2 + 2 r_0^3) \right]_{0}^{R} $$
Get rid of the $-1$ factor by reversing the limits:
$$= 4\pi\rho_c \left[ e^{-\frac{r}{r_0}} (r^2r_0 + 2r r_0^2 + 2 r_0^3) \right]^{0}_{R} $$
Factor out one $r_0$ term:
$$= 4\pi\rho_c r_0 \left[ e^{-\frac{r}{r_0}} (r^2 + 2r r_0 + 2 r_0^3) \right]^{0}_{R} $$
Evaluate bounds, and simplify ($e^0 = 1$, et cetera):
$$m(R) = 4\pi\rho_c r_0 \left[ 2r_0^2 - e^{-\frac{R}{r_0}}(R^2 + 2Rr_0 + 2r_0^2) \right]$$
That is the hard part of the problem done. Consider that the mass can be modelled as being concentrated at the centre of the galaxy. Consider also that, as the other shells of the galaxy further out from the centre have graviational force vectors that all cancel out (i.e. the same reason why the gravitational force in a hollow spherical planet is 0), this is the only mass we need to worry about. You can take Newton's Law of Gravitation and the Circular Motion equation to get:
$$G\frac{m(r)}{r^2} = \mathrm{acceleration} = \frac{v(r)^2}{r}$$
$$v(r) = \sqrt{\frac{Gm(r)}{r}}$$
For your final answer:
$$v(r) = \sqrt{\frac{4 G \pi\rho_c r_0 \left[ 2r_0^2 - e^{-\frac{r}{r_0}}(r^2 + 2rr_0 + 2r_0^2) \right]}{r}}$$
The second question is much simpler. Consider the fact that, for $v(r)$ to be constant, $\frac{m(r)}{r}$ must be constant, so that $v(r)$ is always just some multiple of $G$.
For this to be the case, the mass of each shell must therefore be constant, as to integrate those masses with respect to $r$ would then leave you with some $m_{total}(r) = kr$, and $v(r) = \sqrt{kG}$. Suppose then, that $k$ is defined by $v(r) = \sqrt{kG}$, where $v(r)$ is the observed circumferential velocity. As $m_{total}(r) = kr$, and $m_{total}(r) = m(r) + m_{dark\,matter}(r)$; $m_{dark\,matter}(r) = kr - m(r)$.
$$m_{dark\,matter}(r) = kr - \left( 4\pi\rho_c r_0 \left[ 2r_0^2 - e^{-\frac{r}{r_0}}(r^2 + 2rr_0 + 2r_0^2) \right] \right)$$
...is thus the final answer, using the expression for $m(r)$ we derived to answer the first question.
