Find the length of side PQ using the Pythagorean theorem in the right-angled triangle PQR.

We have $QR = 4$ and $PR = 13$.

$$PQ^2 + QR^2 = PR^2$$ $$PQ^2 + 4^2 = 13^2$$ $$PQ^2 + 16 = 169$$ $$PQ^2 = 169 - 16$$ $$PQ^2 = 153$$ $$PQ = \sqrt{153} = \sqrt{9 \times 17} = 3\sqrt{17}$$

Angle PRT and angle PRQ are supplementary angles, meaning their sum is $180^\circ$. Therefore, $\sin PRT = \sin(180^\circ - PRQ) = \sin PRQ$.
In $\triangle PQR$, $\sin PRQ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{PQ}{PR}$.

$$\sin PRT = \frac{3\sqrt{17}}{13}$$

The value of $\sin PRT$ is $\boxed{\frac{3\sqrt{17}}{13}}$.

In $\triangle PQR$, for angle QPR, PQ is the adjacent side and PR is the hypotenuse.

$$\cos QPR = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PQ}{PR}$$ $$\cos QPR = \frac{3\sqrt{17}}{13}$$

The value of $\cos QPR$ is $\boxed{\frac{3\sqrt{17}}{13}}$.

Angle PRT and angle PRQ are supplementary angles. Therefore, $\cos PRT = \cos(180^\circ - PRQ) = -\cos PRQ$.
In $\triangle PQR$, $\cos PRQ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{QR}{PR}$.

$$\cos PRT = -\frac{QR}{PR} = -\frac{4}{13}$$

The value of $\cos PRT$ is $\boxed{-\frac{4}{13}}$.

In $\triangle PQR$, for angle QPR, QR is the opposite side and PQ is the adjacent side.

$$\tan QPR = \frac{\text{opposite}}{\text{adjacent}} = \frac{QR}{PQ}$$ $$\tan QPR = \frac{4}{3\sqrt{17}}$$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{17}$:

$$\tan QPR = \frac{4\sqrt{17}}{3\sqrt{17} \times \sqrt{17}} = \frac{4\sqrt{17}}{3 \times 17} = \frac{4\sqrt{17}}{51}$$

The value of $\tan QPR$ is $\boxed{\frac{4\sqrt{17}}{51}}$.