Q1: assume that degenerate quadruplets do not exist (they have probability 0 of occuring). Observe that for any non-degenerate quadruple of points $P = (p_1, p_2, p_3, p_4)$ we have exactly one of three options:

- $P$ is a quadrilateral;
- $p_1 p_2$ intersects $p_3 p_4$;
- $p_1 p_4$ intersects $p_2 p_3$.

Let $B$ be the event of segments $p_1 p_2$ and $p_3 p_4$ crossing for random points $p_1, p_2, p_3, p_4$, then the probability of $A = \{P$ that form a quadrilateral$\}$ is $1 - 2 \Pr(B)$.

To find $\Pr(B)$, assign the unique intersection point $p$ to each $(p_{11}, p_{12}, p_{21}, p_{22}) \in B$. Introduce an injective change of coordinates among quadruplets $P$ for which $p$ is defined:

- $(r, \alpha)$ = polar coordinates of $p$;
- $(l_{ij}, \alpha_{ij})$ = polar coordinates of $p_{ij}$ with respect to center $p$ and reference direction $\alpha$.

Since $p_{11} p_{12}, p_{21} p_{22} \ni p$, we must have $\alpha_i := \alpha_{i1} = \alpha_{i2} + \pi$ for $i = 1, 2$. $\Pr(B)$ can be found by integrating Jacobian $J = (l_{11} + l_{12})(l_{21} + l_{22})r \sin(\alpha_1 - \alpha_2)$ (up to a sign) over the region where all points lie within the circle (we only integrate for $0 \leq \alpha_1 \leq \alpha_2 \leq \pi$, but multiply by $8$ to account for different orders):
$\Pr(B) = 8\int_0^1 dr \int_0^{2\pi} d \alpha \int_0^{\pi} d \alpha_1 \int_{\alpha_1}^{\pi} d \alpha_2 \int_0^{\sqrt{1 - r^2 \sin^2 \alpha_1} - r\cos \alpha_1} dl_{11} \int_0^{\sqrt{1 - r^2 \sin^2 \alpha_1} + r\cos \alpha_1} dl_{12} \int_0^{\sqrt{1 - r^2 \sin^2 \alpha_2} - r\cos \alpha_2} dl_{21} \int_0^{\sqrt{1 - r^2 \sin^2 \alpha_2} + r\cos \alpha_2} dl_{22} \cdot (l_{11} + l_{12})(l_{21} + l_{22})r \sin(\alpha_2 - \alpha_1).$

Integrating $J$ within required bounds seems hard (probably no closed answer...), but numerical integration yields $\Pr(B) \approx 0.23482663$, and $\Pr(A) \approx 0.53034674$, which agrees well with my Monte Carlo computations.

Q2: let $C$ be the event "an *unordered* set of four random points is in convex position". Considering all $4!$ orderings in each case we can conclude that $\Pr(A) = \frac{1}{3}\Pr(C) + \Pr(\overline{C})$, which implies $\Pr(C) = \frac{3}{2}(1 - \Pr(A)) \approx 0.70448$. A convex quardilateral is lit by a random interior point with probability 1.

Consider a non-convex quadrilateral with convex hull $H = p_1 p_2 p_3$. Conditioned on vertices of $H$, the fourth point $p_4$ is uniformly distributed within the triangle $H$, and the side of $H$ to which $p_4$ is connected is equidistributed. Notice that affine transformations don't change uniformity of distribution nor that a particular interior point lights the quadrilateral. It follows that conditional probability "a random interior point lights a random non-convex quardilateral with convex hull $H$" does not depend on $H$. Let us find it for a standard triangle with vertices $(0, 0), (1, 0), (0, 1)$: $$\frac{\int_0^1 dx \int_0^{1 - x} dy \frac{1 - y(\frac{x}{1 - x} + \frac{1}{x + y})}{1 - y}}{1/2} = 7 - \frac{2\pi^2}{3} \approx 0.420264$$

Hence $$\Pr(\text{a random quadrilateral is lit by a random interior point}) \approx \frac{\frac{1}{3}\Pr(C) + 0.420264\Pr(\overline{C})}{\Pr{A}} \approx 0.676959,$$ which, again, seems to meet the Monte Carlo values pretty well.