Ans: $\angle RQS=27^\circ$, $\angle PQS=63^\circ$
In $\Delta QRT$,
$\begin{array}{rcll}
\angle QRT & = & \angle PSQ & \text{($\angle$ in the same segment)} \\
\angle QRT & = & 41^\circ \\
\angle RQS & = & \angle PTQ -\angle QRT & \text{(ext. $\angle$ of $\Delta$)} \\
\angle RQS & = & 68^\circ -41^\circ \\
\angle RQS & = & 27^\circ
\end{array}$
Since $PR$ is a diameter of the circle,
$\begin{array}{rcll}
\angle PQR & = & 90^\circ & \text{($\angle$ in semi-circle)}
\end{array}$
Hence, we have
$\begin{array}{rcl}
\angle PQS & = & \angle PQR -\angle RQS \\
\angle PQS & = & 90^\circ- 27^\circ \\
\angle PQS & = & 63^\circ
\end{array}$
