Here are the solutions to the four geometry problems:
1. Since \angle PAQ = 13^\circ and \overline{AP} is an altitude of the triangle, we have \angle PBA = 90^\circ - 13^\circ = 77^\circ. Also, since \angle B > \angle C, we have \angle BAC > \angle BAC. Therefore, \angle RAC = 90^\circ - \angle BAC < 90^\circ - 77^\circ = 13^\circ.
2. Since M, N, and P are the midpoints of sides \overline{TU}, \overline{US}, and \overline{ST} of triangle STU, respectively, we have \overline{MP} = \overline{PN} = \overline{NM}. Also, since \overline{UZ} is an altitude of the triangle, we have \angle NZM = \angle NPM.
Therefore, \angle NZM + \angle NPM = 2 \angle NZM = 2 \angle UZN = 2(90^\circ - 73^\circ) = 34^\circ.
3. Since M, N, and O are the midpoints of sides \overline{KL}, \overline{LJ}, and \overline{JK}, respectively, of triangle JKL, we have \overline{MO} = \overline{ON} = \overline{NP} = \overline{PQ} = \overline{QR}. Also, since P, Q, and R are the midpoints of \overline{NO}, \overline{OM}, and \overline{MN}, respectively, we have \overline{PQ} = \overline{QR} = 2 \overline{PR}.
Therefore, the area of triangle PQR is:
[PQR] = 1/2 * PQ * QR = 1/2 * 2PR * 2PR = 2PR^2
Since the area of triangle PQR is 27, we have:
2PR^2 = 27
PR = \sqrt{27/2} = 3\sqrt{3/2}
Therefore, the area of triangle LPQ is:
[LPQ] = 1/2 * LP * PQ = 1/2 * (2PR) * 2PR = 2PR^2 = 2(3\sqrt{3/2})^2 = 54\sqrt{3}
4. Since D is the midpoint of median \overline{AM} of triangle ABC, we have \overline{MD} = \overline{DA}. Also, since E is the midpoint of \overline{AB}, we have \overline{AE} = \overline{EB}.
Therefore, triangle DMT is congruent to triangle AEB. Since the area of triangle ABC is 180, the area of triangle AEB is 90. Therefore, the area of triangle DMT is also 90.
Summary:
\angle RAC = 13^\circ
\angle NZM + \angle NPM = 34^\circ
[LPQ] = 54\sqrt{3}
[DMT] = 90
