= Solution
Here representations mean finite-dimensional <rational representations> of $SL_2(\mathbb C)$. Put $W=\mathbb C^2$. The irreducibles are
$$
\boxed{V_d=\operatorname{Sym}^dW,\qquad d=0,1,2,\ldots,\qquad\dim V_d=d+1.}
$$
The invariant alternating form on $W$ identifies $W$ with $W^*$, so these can equally be realized as <binary forms> of degree $d$. With basis $e_j=x^{d-j}y^j$, the standard <Lie algebra> generators act as
$$
H=x\partial_x-y\partial_y,\quad E=x\partial_y,\quad F=y\partial_x;\qquad He_j=(d-2j)e_j,\quad Ee_j=je_{j-1},\quad Fe_j=(d-j)e_{j+1}.
$$
The distinct torus weights and these nonzero lowering and raising coefficients show irreducibility: any nonzero <invariant subspace> contains a <weight vector>, and repeated applications of $E,F$ obtain the entire basis.
For exhaustion, choose a <highest-weight vector> $v$ of an irreducible representation, so $Hv=mv$, $Ev=0$. The <sl2 highest-weight lowering formula>, obtained by induction from $[E,F]=H$, is $EF^jv=j(m-j+1)F^{j-1}v$. If $F^rv\ne0$ but $F^{r+1}v=0$, applying $E$ to the latter equality gives $(r+1)(m-r)F^rv=0$, hence $m=r$. The vectors $v,Fv,\ldots,F^rv$ have distinct weights and span an invariant <submodule>, so irreducibility makes it the whole representation. After rescaling basis vectors it is $V_r$. Complete reducibility from Question 2 now describes every representation as a finite direct sum of these <symmetric powers>. Their central element $-I$ acts by $(-1)^d$.
The <symbolic method for binary-form invariants> works as follows. Write a degree-$d$ form as $f(x,y)=\sum_{j=0}^d\binom dj f_jx^{d-j}y^j$ and introduce a formal letter $a=(a_0,a_1)$. The formal power $(a_0x+a_1y)^d$ is converted to $f$ by the linear substitution $a_0^{d-j}a_1^j\mapsto f_j$. This is an umbral linear rule, not a claim that every form is an actual power of a linear form.
For a coefficient polynomial of degree $r$, polarize it into $r$ copies of the form, represented by letters $a,b,\ldots$. Each letter must occur with degree $d$. In <characteristic zero>, polarization and evaluation on equal forms recover the original coefficient polynomial. The basic invariant expressions in the letters are brackets
$$
(ab)=a_0b_1-a_1b_0.
$$
They are invariant because $\det g=1$. <Invariant tensors> on copies of a two-dimensional space are spanned by pairwise alternating contractions; equivalently, the invariant polynomials in the letters are generated by these brackets. One can see the tensor statement from the two-row Young-symmetrizer description of the <determinant> powers in tensor space. Their basic relations include $(ab)=-(ba)$ and $(ab)(cd)-(ac)(bd)+(ad)(bc)=0$. Multiply brackets, ensure the required degree in each letter, symmetrize letter copies as needed, and apply the umbral substitutions. This gives all coefficient invariants, and also works for joint invariants of several forms.
For example, for $f=Ax^2+2Bxy+Cy^2$, the expression $(ab)^2$ becomes $2(AC-B^2)$, the quadratic discriminant up to its conventional scalar. Contractions leaving variable indices unpaired give covariants. A useful systematic construction is the <transvectant of binary forms>:
$$
(f,g)_r=\frac{(m-r)!(n-r)!}{m!n!}\sum_{j=0}^r(-1)^j\binom rj(\partial_x^{r-j}\partial_y^jf)(\partial_x^j\partial_y^{r-j}g).
$$
For forms of degrees $m,n$, this is an equivariant form of degree $m+n-2r$. The corresponding tensor decomposition is
$$
\boxed{\operatorname{Sym}^mW\otimes\operatorname{Sym}^nW\cong\bigoplus_{r=0}^{\min(m,n)}\operatorname{Sym}^{m+n-2r}W.}
$$
Its character identity follows by multiplying the weight sums; complete reducibility then gives the decomposition, and alternating contractions supply the displayed maps onto its summands.
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