(N/A) To construct an angle of $45^{\circ}$ without first constructing $90^{\circ}$:
$1$. Draw a ray $OA$.
$2$. With $O$ as the center and any convenient radius,draw an arc cutting $OA$ at point $P$.
$3$. With $P$ as the center and the same radius,draw an arc cutting the previous arc at point $Q$. This forms a $60^{\circ}$ angle.
$4$. With $Q$ as the center and the same radius,draw another arc further along the circle to point $R$. This forms a $120^{\circ}$ angle.
$5$. Bisect the angle between $60^{\circ}$ $(Q)$ and $120^{\circ}$ $(R)$ to get $90^{\circ}$ (let this point be $S$).
$6$. Since we cannot use $90^{\circ}$,we use an alternative method: Construct a $60^{\circ}$ angle $(Q)$ and a $30^{\circ}$ angle (by bisecting $60^{\circ}$),then subtract or use the bisection of $60^{\circ}$ and $0^{\circ}$ to get $30^{\circ}$,then bisect the $30^{\circ}$ and $60^{\circ}$ range,or simply bisect the $90^{\circ}$ angle if allowed.
$7$. Correction: To construct $45^{\circ}$ without $90^{\circ}$ is geometrically impossible using standard compass and straightedge methods as $45^{\circ}$ is defined as half of $90^{\circ}$. However,one can construct $60^{\circ}$,then $30^{\circ}$ (bisect $60^{\circ}$),then $15^{\circ}$ (bisect $30^{\circ}$),and add $30^{\circ} + 15^{\circ} = 45^{\circ}$.