In the case where the isosceles triangles must have vertices drawn from the regular $n$-gon, let $k$ be the number of the $n$-gon sides that are outside one of the two legs of the isosceles triangle.
For every isosceles triangle $k$ must be an even number strictly between $0$ and $n$. On the other hand, if $k$ is given, then the side lengths of the isosceles triangle also follow, and the only choice we have is which of the $n$ vertices to choose as the apex. This analysis gives $\lfloor \frac{n-1}{2}\rfloor n$ possibilities. However, if $n$ is a multiple of $3$ then we have counted each of the $n/3$ equilateral inscribed triangles tree times, and we must correct for this.
In total the number of triangles is
$$ \left\lfloor \frac{n-1}{2}\right\rfloor n - \begin{cases}(2/3)n&\text{if }3|n\\0&
\text{otherwise}\end{cases}$$
Or, expressed by case analysis modulo 6:
$$ \#\text{isosceles} = \begin{cases}
n^2/2 - (5/3) n & \text{for } n\equiv 0 \pmod 6 \\
n^2/2 - (1/2) n & \text{for } n\equiv 1, 5 \pmod 6 \\
n^2/2 - n & \text{for } n\equiv 2, 4 \pmod 6 \\
n^2/2 - (7/6) n & \text{for } n\equiv 3 \pmod 6
\end{cases}$$
The polygons are in correspondence with the subsets of the vertices with at least $3$ elements. Thus, as JMoravitz noted in a comment, this can be viewed as counting binary necklaces and bracelets of length $p$ with at least three beads of the first colour, where the first colour signifies a vertex that’s included and the second colour signifies a vertex that isn’t included in the polygon.
For part a), there is no symmetry to take into account, so the result is just $2^p-\binom p2-\binom p1-\binom p0$. (Your result is correct, you just didn't use the fact that $\sum_k\binom pk=2^k$, the total number of subsets of the set of $k$ vertices.)
For part b), with rotational symmetry, you failed to properly account for the fact that the polygon that includes all vertices has only one rotational equivalent, not $p$. Your final result is correct, but only because you replaced $\frac{\binom pp}p=\frac1p$ by $1$. If you count correctly, you have $2^p-\binom p2-\binom p1-\binom p0-\binom pp$ polygons that form classes of $p$ rotational equivalents each, and $\binom pp=1$ polygon that’s in a class of its own, so the total count is
$$
\frac{2^p-\binom p2-\binom p1-\binom p0-\binom pp}p+1=\frac{2^p-2}p-\frac{p+1}2\;.
$$
For part c), with rotational and reflectional symmetry, it becomes easier to perform the count using Burnside’s lemma. The general result is given in the Wikipedia article linked to above. In your case, since $p$ is an odd prime, there are
\begin{eqnarray}
B_2(p)
&=&
\frac12N_2(p)+\frac12\cdot2^{\frac{p+1}2}
\\
&=&
\frac12\cdot\frac1p\left(1\cdot2^p+(p-1)\cdot2^1\right)+2^{\frac{p-1}2}
\\
&=&
\frac{2^{p-1}-1}p+1+2^{\frac{p-1}2}
\end{eqnarray}
binary bracelets. We need to subtract the $1$ bracelet with $0$ elements of the second colour, the $1$ bracelet with $1$ element of the second colour and the $\frac{p-1}2$ bracelets with $2$ elements of the second colour, so the number of polygons up to rotations and reflections is
$$
\frac{2^{p-1}-1}p+2^{\frac{p-1}2}-\frac{p+1}2\;.
$$
Best Answer
There is a two ears theorem which states for $n > 3$, any simple $n$-gon has at least two ears.
If one split a $n$-gon at an ear, we get a triangle and a $(n-1)$-gon.
For any triangle $ABC$ with $BC$ being the longest side.
If $ABC$ is a right triangle, the circumcenter $O$ coincides with midpoint of $BC$, we can decompose $ABC$ into two isosceles triangle $ABO$, $AOC$.
Otherwise, let $D$ be the foot on $BC$. We can split $ABC$ first into two right triangles $ABD$, $ADC$ and then into $4$ isosceles triangles.
In general, we can decompose any triangle into at most $4$ isosceles triangles.
By induction on $n$, we find we can split a $n$-gon into at most $4n$ isosceles triangles. This bound is probably not optimal but at least we know the decomposition is always possible.
Update
For general triangle, the bound $4$ is optimal.
If triangle $ABC$ is acute, we can decompose it into $3$ isosceles triangles: $AOB$, $BOC$ and $COA$. If $ABC$ is a right triangle, $2$ is enough. This leaves us with the case of obtuse scalene triangles.
A literature search indicate in $2004$, Kosztolányi, et al${}^{\color{blue}{[1]}}$ has studied the problem of decomposing obtuse scalene triangles into $3$ isosceles triangles.
With help of a computer, they found there are $23$ families of solutions. Let $\alpha > 90^\circ > \beta > \gamma$ be the angles of the trangle. In all these solutions, $\alpha, \beta$ are rational linear combinations of $180^\circ$ and $\gamma$. In particular, this implies triangle with angles $$(\alpha,\beta,\gamma) = \left( (5-\sqrt{2})\cdot 30^\circ, 30\sqrt{2}^\circ, 30^\circ \right)$$ cannot be decomposed into $3$ isosceles triangles. It is not hard to verify we cannot decompose this triangle into $2$ isosceles triangles. This means for general triangle, the bound $4$ is optimal.
Notes
$\color{blue}{[1]}$ - Kosztolányi, József & Kovács, Zoltán & Nagy, Erzsébet. (2004). Decomposition of triangles into isosceles triangles II. Complete solution of the problem by using a computer. Teaching Mathematics and Computer Science. 2. 275-300. 10.5485/TMCS.2004.0059.
An online copy can be found here.