This answer focuses on part (C) of the question. See also the excellent answer by @Shyos.
Let's write E for Ebrecedebre, I for Ibricidibri, and similarly for O and U. Then we'll write e, i, o, u for the "reversed" operations. Following @Shyos, we'll also write B for Ball, C for Cone, R for Ring and J for Umbrella, so that we don't have U meaning two different things. (J is an excellent letter for an Umbrella!)
Each time you do E, it decreases the number of B's by 1, while increasing the number of C's by 1 and the number of R's by 1. On the other hand, if you do an "e", it increases the number of B's by 1 while decreasing C's and R's by 1. Notice that E and e just cancel each other out.
Also, if at any point we did, say, 8 E's followed by 5 e's, the net result would be the same as doing (8-5) = 3 instances of E. So we can count the E's as positive, and then each e is just a negative E.
In part (C), we need a total net increase in the R's by at least 5, and we need a total net decrease in the C's by exactly 1, and no net change in the B's or J's.
Let's write $x,y, z, w$ to mean the net number of applications of E, I, O, U respectively. (Note: since e, i, o, u count as negative, it's possible that $x,y,z,w$ could be negative; however, $x,y,z,w$ must all be integers.) Now the total net change in B's is:
$$
\Delta B = -x + 3z + w
$$
(Why? Well, remember that each application of E is going to decrease the number of B's by 1. Meanwhile, each application of O will increase the number of B's by 3, and each application of U will increase the number of B's by 1. So 3 times the number of O's, plus 1 times the number of U's, minus 1 times the number of E's, gives the total net change in the number of B's.)
We want $\Delta B = 0$, and $\Delta C = -1$ (since we start with 1 cone and want to end with 0), and similarly $\Delta J =0$. Finally, $\Delta R$ can be any number $k$ as long as $k\geq 5$. We now have:
$$
\begin{eqnarray*}
0 & = & -x + 0y + 3z + w\\
-1 & = & x - y + 0z + 4w\\
k & = & x + 2y - z + 0w\\
0 & = & 0x + y + z - w\\
\end{eqnarray*}
$$
Or in matrix form:
$$
\begin{pmatrix}
0\\
-1\\
k \\
0 \\
\end{pmatrix}
=
\begin{pmatrix}
-1 & 0 & 3 & 1\\
1 & -1 & 0 & 4\\
1 & 2 & -1 & 0\\
0 & 1 & 1 & -1\\
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z \\
w \\
\end{pmatrix}
$$
We can solve this equation for $x,y,z,w$ in a variety of ways; for example, multiplying both sides by the inverse of the $4\times 4$ matrix. No matter how you do it, the result is:
$$
\begin{pmatrix}
x\\
y\\
z \\
w \\
\end{pmatrix}
=
\begin{pmatrix}
-{3}/{4} - 2k/3\\
{1}/{4} + {2k}/{3}\\
-1/4 - k/3 \\
k/3 \\
\end{pmatrix}
$$
Since $x,y,z,w$ have to be integers, we see that $k$ must be divisible by 3 (from the last entry). Also, from the first entry, the following must be an integer:
$$
-\frac{3}{4} - \frac{2k}{3} = \frac{-9 - 8k}{12},
$$
so $(8k+9)$ must be divisible by $12$. This is impossible if $k$ is divisible by $3$, so there is no solution to part (C).
