This was a really nice question thanks for posting it @Alireza_Amiri Throughout the solution let T denote the given foot of the altitude and let F denote the given midpoint of the arc. a) First construct line PT, let this be e. Let the intersection of the line through F parallel with e and the line through T perpendicular to e be D. Then point D is the midpoint of side BC. Reflect P by T and D, let these points be P′ and P′′ respectively. It is known that P′ and P′′ lie on the circumscribed circle of ABC, this can be verified by some angle chasing. P′, F and P′′ are pairwise different since ABC is scalene, so by constructing the circle P′FP′′ we get the circumscribed circle of ABC, intersecting it with lines PT and TD we get points A, P′, B and C. b) We will use the following fact multiple times: the centroid divides the medians in the ratio of 2 : 1. It follows from this fact that point T′, which we get by enlarging point T from P with scale factor −2, lies on the circumcircle of ABC and coincides with the the reflection of point A by the perpendicular bisector of side BC. It also follows that F′, which we get by enlarging point F from point P with scale factor −2, lies on line AT. Now let us start the construction. We can construct ′′′ points F and T . Consider the line that is the enlargement of line TF from P with scale factor −1/2. This will be the perpendicular bisector of side BC; by reflecting T′ by this line, we get A. Since triangle ABC is scalene, this means that A ̸= T′ so we can construct the circumscribed circle of ABC, knowing its three different points. By intersecting this circle with the line perpendicular to TF′ through T, we get B and C. c) A few remarks. Let the circumcircle of triangle ABC be k1. It is known that points B, P and C lie on a circle with center F let this be k2 (this can be verified by angle chasing). Let the intersection of lines AF and BC be M. Let the circle with diameter AM be k3, denote its center with O. Since T is the foot of the altitude through A, triangle ATM is right, this means that T lies on k3. Let T′ denote the second intersection of line FT with k3. Since the inversion with respect to k2 maps k1 to line BC, so the image of pont M is point A, so the image of k3 is itself, it means that the image of T ′ is T . The construction: (we are using the same notations as in the remarks). First construct k2 (knowing its center and one point on the perimeter). Construct T′, the image of T in the inversion with respect to k2. Now the intersection of line FP and the perpendicular bisector of TT′ is O, so we can construct k3. The intersection of k3 and line FP farther from F will be A. The intersections of k2 and the line through T perpendicular to AT will be points B and C.

Alternative solution for (c): Let $D$ be the foot of the A-altitude, $S$ be the arc midpoint and $I$ the incenter. First the construction: Let $I_A$ be the A-excircle of $ABC$ which we construct by reflecting $I$ over $S$ Next we draw $l$, the angle bisector of $\angle I_ADI$, this is the line $BC$ From here we can find B,C by intersecting $l$ with the circle centered at $S$ through $I$. Then we may find $A$ by intersectingthe circle $BCS$ with $IS$. As for why the angle bisector is true, consider the following: in triangle $ABC$ there is a homothety centered at $A$ which sends the incircle to the A-excircle. This sends $D$ to $D'$ and the line $BC$ to a line parallel to $BC$ tangent to the excircle Thus $I_A$ must lie on the perpendicular bisector of $DD'$ thus we have $\angle IDD' = \angle ID'D$ from which the conclusion follows, since the homothety sends $I$ to $I'$