VIII OBM

For the first part there are three cases to consider: $L$ a semicircle and $P$ inside it, $L$ a semicircle and $P$ outside it, and $L$ a perpendicular half-line. The diagram above shows the first case. Take $L$ to have radius $R$. The semicircle through $P$ with the same center $O$ as $L$ does not intersect $L$. Suppose its radius is $R-k$. Now take a semicircle through $P$ with center $X$ such that $OX<\frac{k}{2}$. Then its radius $XP<OP+OX<R-\frac{k}{2}$, and $\min (XA,XB)>R-\frac{k}{2}$, so the semicircle lies entirely inside $L$. Thus we have an infinity of possible semicircles through $P$.

An exactly similar argument works for $P$ outside $L$.

For the third case, just take the center sufficiently far from $P$ on the same side of $L$ as $P$.

On to the second part. Step 1 is to show that if we fix $C$ and allow $B$ to vary along a line $l$, then $\angle x$ increases as $B$ moves down ($O$ is confined to the line and $OB=OC$). In fact, if $M$ is the midpoint of $BC$ then $M$ belongs to the line parallel to $l$ and passes through a point whose distance to $l$ is half the distance from $C$ to $l$. Moreover, the quadrilateral $OPBM$ is cyclic, so $\angle OBM=\angle OPM$, and $\angle OPM$ decreases as $B$ moves down. This means that $\angle x=90^{\circ }-\angle OPM$ increases as $B$ moves down.

Step 2 is to show that $\angle b+\angle c+\angle x=180^{\circ }$. Let the tangents to the arc passing through $B$ and $C$ intersect in $D$. So $\angle BDC=\angle b+\angle c$ and, by looking at the internal angles of the quadrilateral $OBDC$, $\angle x+90^{\circ }+90^{\circ }+\angle b+\angle c=360^{\circ }⟺\angle b+\angle c+\angle x=180^{\circ }$.

Now consider the special case where $AB$ is a straight line. Using the previous results one can show by some angle chasing that the sum of the angles in the triangle is $180^{\circ }-(\angle O^{\prime }-\angle O)<180^{\circ }$. Finally, use this special case to deduce the other cases (by dividing the general triangle into two parts which fall into the special case).