Identify the angle in a semicircle.

Since QOS is a diameter, the angle subtended by the diameter at any point on the circumference is $90^\circ$.
Therefore, $\angle QRS = 90^\circ$.

The angle between the tangent ST and the chord SQ at the point of contact S is $\angle QST$. According to the Tangent-Chord Theorem, this angle is equal to the angle in the alternate segment, which is $\angle SRQ$.
So, $\angle QST = \angle SRQ$.
From Step 1, we know $\angle SRQ = \angle QRS = 90^\circ$.
Therefore, $\angle QST = 90^\circ$.

We are given $\angle QST = 3x + 15^\circ$. Equating this to $90^\circ$:

The value of $x$ is $\boxed{\text{}25^\circ\text{}}$.

Consider $\triangle QRS$. The sum of angles in a triangle is $180^\circ$.
We know:

• $\angle QRS = 90^\circ$ (from Step 1)
• $\angle RQS = x^\circ = 25^\circ$ (given and from Step 3)

The value of $\angle RSQ$ is $\boxed{\text{}65^\circ\text{}}$.