Consider the spherical right triangle $NPQ$, where $N$ is the north pole, $P$ is the starting location, and $Q$ is the position after sailing $1000$ miles. For simplicity let the radius of the earth be $1$, so that arclengths are equal to the angle they subtend.

The latitude of $Q$ is given by $NQ$, and the difference in longitudes between $P$ and $Q$ is given by $\angle PNQ$. The direction that the ship is traveling at $Q$ can be described by using the angle of the ships velocity vector with the longitude line it currently is at.

In other words, the ship will be traveling at $\angle PQN$ degrees East of South.

It remains to solve the right spherical triangle. We are given that $NP = 0.780$ radians, and $PQ = 0.291$ radians. Then by the standard formula for right spherical triangles, one obtains:

$$NQ = \cos^{-1}(\cos(NP)\cos(PQ))$$

$$\angle PNQ = \cos^{-1}(\tan(PN)\cot(NQ))$$

$$\angle PQN = \cos^{-1}(\tan(PQ)\cot(QN))$$

This implies that the latitude of the ship is now $47.08^\circ$ and the longitude is $63.58^\circ + 23.07^\circ = 86.65^\circ$. The direction the ship is traveling is furthermore $73.83^\circ$ East of South.