Why is dU an exact differential and dq an inexact differential?
As opposed to an exact differential, an inexact differential cannot be expressed as the differential of a function, i.e. while there exist a function $U$ such that $U = \int \mathrm{d} U$, there is no such functions for $\text{đ} q$ and $\text{đ} w$.
And the same is, of course, true for any state function $a$ and any path function $b$ respectively: an infinitesimal change in a state function is represented by an exact differential $\mathrm{d} a$ and there is a function $a$ such that $a = \int \mathrm{d} a$, while an infinitesimal change in a path function $b$ is represented by an inexact differential $\text{đ} b$ and there is no function $b$ such that $b = \int \text{đ} b$.

Consequently, for a process in which a system goes from state $1$ to state $2$ a change in a state function $a$ can be evaluated simply as
$$\int_{1}^{2} \mathrm{d} a = a_{2} - a_{1} \, ,$$ 
while a change in a path function $b$ can not be evaluated in such a simple way,
$$\int_{1}^{2} \text{đ} b \neq b_{2} - b_{1} \, .$$
And for a state function $a$ in a thermodynamic cycle
$$\oint \mathrm{d} a = 0 \, ,$$
while for a path function $b$
$$\oint \text{đ} b \neq 0 \, .$$
The last mathematical relations are important, for instance, for the first law of thermodynamics, because while $\oint \text{đ} q \neq 0$ and $\oint \text{đ} w \neq 0$ it was experimentally found that $\oint (\text{đ} q + \text{đ} w) = 0$ for a closed system, which implies that there exist a state function $U$ such that $\mathrm{d} U = \text{đ} q + \text{đ} w$.