= Solution
With the right-<comodule> <braiding> convention in part (a), the defining formula is
$$
C(e_i\otimes e_j)=\sum_{k,l}e_l\otimes e_k\,r(t_{ki},t_{lj}).
$$
Consequently the scalar form on generators is the following table, with row the first argument and column the second:
$$
\boxed{\begin{array}{c|rrrr}
r&a&b&c&d\\\hline
a&q^{1/2}&0&0&q^{-1/2}\\
b&0&0&q^{-1/2}(q-q^{-1})&0\\
c&0&0&0&0\\
d&q^{-1/2}&0&0&q^{1/2}
\end{array}.}
$$
For example, the extra $e_2\otimes e_1$ component of $C(e_2\otimes e_1)$ is $r(b,c)=q^{-1/2}(q-q^{-1})$. The other off-diagonal component gives $r(d,a)=q^{-1/2}$.
For arbitrary elements of the <FRT bialgebra>, extend this table by bilinearity, the unit rule $r(1,z)=r(z,1)=\varepsilon(z)$, and
$$
r(uv,z)=\sum r(u,z_{(1)})r(v,z_{(2)}),\qquad
r(u,vw)=\sum r(u_{(1)},w)r(u_{(2)},v).
$$
These are an explicit recursive algorithm for any two words in the generators, so the table determines the entire <cobraiding>. The braid equation for $C$ ensures the algorithm annihilates the defining coefficient relations in either argument: moving one crossing past two successive crossings gives precisely the two reductions of a quadratic colinearity relation. Thus it descends from the free <algebra over a field> to the <FRT bialgebra>. For the convolution inverse, use the inverse of $R=\tau C$, where $\tau$ flips the tensor factors: its coefficients are $\bar r(t_{ki},t_{lj})=(R^{-1})_{kl,ij}$. Equivalently, in the flipped coefficient formula use $\tau C^{-1}\tau$, not $C^{-1}$ alone. This gives $\bar r(a,a)=\bar r(d,d)=q^{-1/2}$, $\bar r(a,d)=\bar r(d,a)=q^{1/2}$, $\bar r(b,c)=-q^{1/2}(q-q^{-1})$, and all other generator values zero. This specifies the <coquasitriangular structure> completely, not just its values on degree-one elements.
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