Throughout the solution the following notations will be used:

$C_1(a, b)$: The smaller part of the perimeter of the polygon which lies between $a$ and $b$.

$C_2(a, b)$: The other part of the perimeter of the polygon joining $a$ and $b$.

$d(a, b)$: The length of $C_1(a, b)$.

$[a, b]$: A chord of the polygon whose end points are $a$ and $b$.

a.
Let $[x, y]$ be the maximum chord of $A_o$ with unit length. Our aim is to show that in this case the semicircle with diameter $[x, y]$, which is on the same side of $[x, y]$ as $C_2(x, y)$, lies entirely inside $A_o$.

Lemma 1. $C_2(x, y) \cap [x, y] = \{x, y\}$.

Proof. Assume that $z \in C_2(x, y) \cap ([x, y] - \{x, y\})$. Obviously $[x, z]$ and $[y, z]$ are both inner chords of $A_o$. Note that if $y \in C_1(x, z)$, then $d(x, z) > d(x, y)$ which is a contradiction since chord $[x, y]$ is maximal, hence $y \notin C_1(x, z)$. Similarly, $x \notin C_1(y, z)$. Therefore, $C_1(x, y)$, $C_1(y, z)$ and $C_1(z, x)$ form a partition of the perimeter of $A_o$. This leads to a contradiction because $A_o$ is rotund and $d(x, y)$, $d(y, z)$ and $d(z, x)$ are all less than $\frac{p}{4}$ and consequently their sum is less than $p$. $\square$

Lemma 2. The common part of $C_2(x, y)$ and edges of $A_o$ that contain $x$ lies outside the semicircle mentioned above. A similar assertion holds for $y$.

Proof. Assume to the contrary that $z$ is a point inside that semicircle, and it also lies on the common part of $C_2(x, y)$ and the edge containing $x$ such that $d(x, z)$ is very small. By lemma 1, we can choose $z$ such that $[y, z]$ intersects the perimeter of $A_o$ only at $y$ and $z$. Since $d(x, y) \le \frac{p}{4}$ and $z$ is near to $x$, $yxz = C_1(y, z)$ which is a contradiction with the maximality of $[x, y]$. $\square$

Lemma 3. $x$ and $y$ are the only intersection points of $C_2(x, y)$ and the semicircle.

Proof. Assume to the contrary that the intersection set is not empty. Let $z$ be a point in this set which has the minimum distance to the segment $[x, y]$. Note that by the above lemmas we know that this minimum is positive. We claim that $[x, z]$ and $[y, z]$ are both inner chords of $A_o$, because if not, there is another point of the polygon inside triangle $xyz$, which contradicts the minimality of $z$.
Rest of the proof is similar to that of lemma 1. $y \notin C_1(x, z)$ and $x \notin C_1(y, z)$, so $C_1(x, y)$, $C_1(y, z)$ and $C_1(z, x)$ form a partition of the perimeter of $A_o$, but we know that the length of each one is at most $\frac{p}{4}$. $\square$

Using lemma 3, $C_2(x, y)$ does not intersect the semicircle (except at $x$ and $y$). Therefore, $C_1(x, y)$ does not intersect the semicircle because the polygon is not self-intersecting. Hence the semicircle fits completely inside $A_o$.

b.
Let $[x, y]$ be the maximum chord. Draw semicircles of radius $1$ and centers at $x$ and $y$ such that they are on the same side of line $xy$ as $C_2(x, y)$. We claim that the common part of these two semicircles (say $S$) fits completely inside $A_o$.

Lemma 4. $C_2(x, y) \cap [x, y] = \{x, y\}$.

Proof. The proof is the same as that one given in part a. $\square$

Lemma 5. The common part of $C_2(x, y)$ and edges of $A_o$ which contain $x$ lies outside of $S$. A similar assertion holds for $y$.

Proof. The proof is similar to the proof of lemma 2.
Assume to the contrary that $z \in S$ is a point in the common part of $C_2(x, y)$ and the edge containing $x$ such that $d(x, z)$ is very small. By lemma 4, we can choose $z$ such that $[y, z]$ intersects the perimeter of $A_o$ exactly at $y$ and $z$. Since $d(x, y) \le \frac{p}{4}$ and $z$ is near $x$, $yxz = C_1(y, z)$ which contradicts the maximality of $[x, y]$. $\square$

Lemma 6. $C_2 \cap S = \{x, y\}$.

Proof. Assume to the contrary that there are points other than $x$ and $y$ in this intersection. Let $z$ be a point in this intersection which has the minimum distance from the segment $[x, y]$. By the above lemmas, we know that this minimum is positive. We claim that $[x, z]$ and $[y, z]$ are both inner chords of $A_o$, because if not, there exists another point of the polygon inside triangle $xyz$ which contradicts the minimality of $z$. So $[x, z]$ and $[y, z]$ are both inner chords of $A_o$. But $z$ lies in $S$, therefore the lengths of $[x, z]$ and $[y, z]$ are less than $1$. This implies that $d(x, z)$ and $d(y, z)$ are both less than $\frac{p}{4}$.
Rest of the proof is similar to lemma 1. $y \notin C_1(x, z)$ and $x \notin C_1(y, z)$, so $C_1(x, y)$, $C_1(y, z)$ and $C_1(z, x)$ form a partition of the perimeter of $A_o$, but we know that the length of each one is at most $\frac{p}{4}$. $\square$

By lemma 6, $C_2(x, y)$ does not intersect the region $S$ (except at $x$ and $y$). Therefore, $C_1(x, y)$ does not intersect $S$ because the polygon is not self-intersecting. Hence the region $S$ fits completely inside $A_o$. It is easy to see that a circle with radius $\frac{1}{4}$ can be drawn entirely inside $S$.

c.
Suppose that there exists a rotund polygon $A$ such that no circle of radius $\frac{1}{4}$ completely fits inside it. By the second statement, there is a triangulation of the polygon with chords of at most unit length. Let $xy$ be the chord of the triangulation that has the maximum $d(x, y)$. There are two possibilities:

Case 1. There is no vertex of $A$ in $C_2(x, y)$ other than $x$ and $y$. Therefore, $[x, y]$ must be an edge of $A$. Hence $C_2(x, y) = [x, y]$ and the length of $C_2(x, y)$ is less than $C_1(x, y)$ because of the triangle inequality which is a contradiction.

Case 2. There is another vertex of $A$ in $C_2(x, y)$ other than $x$ and $y$. Therefore, there is some vertex $z \in C_2(x, y)$ such that $xyz$ forms one of the triangles of the triangulation. Note that if $y \in C_1(x, z)$, then $d(x, z) > d(x, y)$ and this contradicts the maximality of $[x, y]$. Thus $y \in C_2(x, z)$ and similarly $z \in C_2(x, y)$. Therefore, $C_1(x, y)$, $C_2(x, y)$ and $C_3(x, y)$ form a partition of the perimeter of $A$. This is a contradiction because the lengths of all these parts are less than $\frac{p}{4}$.

d.
If $\epsilon$ is sufficiently small, the following polygon is a counterexample for the first statement.

e.
[No specific answer provided.]