(a) Three lines $l,m,n$ in space pass through point $S$. A plane perpendicular to $m$ intersects $l,m,n $ at $A,B,C$ respectively. Suppose that $\angle ASB = \angle BSC = 45^o$ and $\angle ABC = 90^o$. Compute $\angle ASC$.
(b) Furthermore, if a plane perpendicular to $l$ intersects  $l,m,n$  at $P,Q,R$ respectively and $SP = 1$, find the sides of triangle $PQR$.
### Part (a)

1. **Understanding the Geometry:**
   - We are given three lines \( l, m, n \) passing through a common point \( S \).
   - A plane perpendicular to \( m \) intersects \( l, m, n \) at points \( A, B, C \) respectively.
   - Given angles: \( \angle ASB = \angle BSC = 45^\circ \) and \( \angle ABC = 90^\circ \).

2. **Visualizing the Cube:**
   - The problem's conditions suggest a cube where \( S \) is a vertex.
   - \( m \) is an edge of the cube containing \( S \) and \( B \).
   - \( l \) and \( n \) are diagonals of the faces of the cube that share the edge \( BS \).
   - Points \( A, B, C \) lie on a single face of the cube.

3. **Equilateral Triangle:**
   - Since \( \angle ABC = 90^\circ \) and \( \angle ASB = \angle BSC = 45^\circ \), \( \triangle ABC \) forms a right triangle with \( \angle ABC = 90^\circ \).
   - The diagonals of the cube's face are equal, making \( \triangle ASC \) an equilateral triangle.
   - Therefore, \( \angle ASC = 60^\circ \).

\[
\boxed{60^\circ}
\]

### Part (b)

1. **New Cube Setup:**
   - Consider a new cube with \( S \) as a vertex.
   - A plane perpendicular to \( l \) intersects \( l, m, n \) at points \( P, Q, R \) respectively.
   - Given \( SP = 1 \).

2. **Right Triangles:**
   - The plane perpendicular to \( l \) cuts through the cube, forming right triangles \( \triangle PQS \) and \( \triangle PQR \).
   - Since \( SP = 1 \), \( PQ \) is also 1 (as it is an edge of the cube).

3. **Applying the Pythagorean Theorem:**
   - In \( \triangle PQS \), since \( PQ = 1 \) and \( SP = 1 \), \( SQ \) can be found using the Pythagorean theorem:
     \[
     SQ = \sqrt{SP^2 + PQ^2} = \sqrt{1^2 + 1^2} = \sqrt{2}
     \]
   - Similarly, in \( \triangle PQR \), since \( PQ = 1 \) and \( QR = \sqrt{2} \), \( PR \) can be found using the Pythagorean theorem:
     \[
     PR = \sqrt{PQ^2 + QR^2} = \sqrt{1^2 + (\sqrt{2})^2} = \sqrt{1 + 2} = \sqrt{3}
     \]

\[
\boxed{PQ = 1, QR = \sqrt{2}, PR = \sqrt{3}}
\]