= Solution
A <Hopf algebra> over $k$ is a unital associative algebra $H$ equipped with a <coalgebra> structure $(\Delta,\varepsilon)$ such that both maps are algebra homomorphisms, and with an <antipode> $S:H\to H$. The defining identities are
$$
(\Delta\otimes1)\Delta=(1\otimes\Delta)\Delta,\qquad
(\varepsilon\otimes1)\Delta=1=(1\otimes\varepsilon)\Delta,
$$
$$
m(S\otimes1)\Delta=\eta\varepsilon=m(1\otimes S)\Delta,
$$
where $m$ is multiplication and $\eta:k\to H$ is the unit map; the counit identities use the standard identifications with $k\otimes H$ and $H\otimes k$. The first compatibility conditions define a <bialgebra>; the <antipode> provides its convolution inverse. Multiplication commutativity and coproduct cocommutativity are independent conditions, as the two examples below demonstrate.
For the final construction, let $q\in\mathbb C^\times$, with $q\ne\pm1$, and take the numerical <Yang-Baxter R-matrix>, in the ordered basis $11,12,21,22$,
$$
R=\begin{pmatrix}q&0&0&0\\0&1&0&0\\0&q-q^{-1}&1&0\\0&0&0&q\end{pmatrix}.
$$
It is invertible. Its four action rules are $11\mapsto q11$, $12\mapsto12+(q-q^{-1})21$, $21\mapsto21$, $22\mapsto q22$. Applying these rules on the eight basis vectors of $V^{\otimes3}$ verifies $R_{12}R_{13}R_{23}=R_{23}R_{13}R_{12}$, the <Yang-Baxter equation>.
In the <RTT construction>, introduce $T=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ and impose $RT_1T_2=T_2T_1R$, where $T_1=T\otimes I$, $T_2=I\otimes T$. Entrywise multiplication gives
$$
ab=qba,\quad ac=qca,\quad bd=qdb,\quad cd=qdc,\quad bc=cb,\quad
ad-da=(q-q^{-1})bc.
$$
For example its $(1,2)$ entry gives $qab=ba+(q-q^{-1})ab$, hence $ab=qba$. Define
$$
\Delta(t_{ij})=\sum_kt_{ik}\otimes t_{kj},\qquad
\varepsilon(t_{ij})=\delta_{ij}.
$$
These maps descend to the RTT algebra. Indeed two matrices $T',T''$ in commuting tensor factors each satisfy RTT, and
$$
R(T'T'')_1(T'T'')_2
=RT'_1T'_2T''_1T''_2
=T'_2T'_1T''_2T''_1R
=(T'T'')_2(T'T'')_1R.
$$
Matrix associativity gives coassociativity and the identity matrix gives the counit. Thus RTT first gives a <bialgebra>; the <R-matrix> alone does not automatically provide an <antipode>.
Here the <quantum determinant> $D=ad-qbc$ is central and satisfies $\Delta D=D\otimes D$, $\varepsilon D=1$, as follows by substitution of the displayed relations. Impose $D=1$. This quotient is the <coordinate Hopf algebra of quantum SL2>, with
$$
\boxed{S(T)=\begin{pmatrix}d&-q^{-1}b\\-qc&a\end{pmatrix}.}
$$
These assignments extend as an anti-algebra map: for example reversing $ab-qba$ gives $-q^{-1}bd+db=0$, and reversing the determinant relation gives $ad-qcb=1$. The remaining relations are checked in the same way. Direct multiplication gives $TS(T)=S(T)T=I$: the first diagonal entry of $TS(T)$ is $ad-qbc=1$, its second is $da-q^{-1}cb=1$, and the off-diagonal entries vanish by the $q$-commutation relations. These are precisely the <antipode> identities on generators, proving that the quotient is a <Hopf algebra>.
To verify both required failures of symmetry, map this algebra onto the <quantum Laurent plane> with generators $u,u^{-1},v$ and relation $uv=qvu$, by $a\mapsto u$, $b\mapsto v$, $c\mapsto0$, $d\mapsto u^{-1}$. Construct this algebra directly on the <vector space> with basis $u^iv^j$, $i\in\mathbb Z$, $j\ge0$, and product $(u^iv^j)(u^kv^\ell)=q^{-jk}u^{i+k}v^{j+\ell}$. The exponent identity $-jk-(j+\ell)m=-\ell m-j(k+m)$ makes multiplication associative, and the basis products give $uv=qvu$ with $u$ invertible. Since $uv\ne vu$ for $q\ne1$, the <Hopf algebra> is noncommutative. Under the same quotient in both tensor factors,
$$
\Delta b-\Delta^{\rm op}b\longmapsto
u\otimes v+v\otimes u^{-1}-v\otimes u-u^{-1}\otimes v\ne0,
$$
because the four terms are distinct tensor-basis vectors. Hence it is also noncocommutative. \b[The construction yields $\boxed{\mathcal O_q(\operatorname{SL}_2)\text{ noncommutative and noncocommutative}}$, for example at $q=2$.]
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