= Solution
In the degree-completed symmetric-function algebra, the <Cauchy identity for symmetric functions> is
$$
\boxed{\prod_{i,j}(1-x_iy_j)^{-1}
=\sum_{\lambda\in\operatorname{Par}}s_\lambda(x)s_\lambda(y).}
$$
It is a formal identity, so each homogeneous component is finite in the partition index; analytic convergence is not being assumed.
Expand each geometric factor and index the product by finite-support nonnegative <matrices>:
$$
\prod_{i,j}(1-x_iy_j)^{-1}
=\sum_A\prod_i x_i^{\sum_j a_{ij}}
\prod_j y_j^{\sum_i a_{ij}}.
$$
The <RSK correspondence> sends $A$ to a pair of <Semistandard Young tableaux> of equal shape. Its content identities turn the <matrix> weight into the product of the recording tableau's $x$ weight and the insertion tableau's $y$ weight. Summing independently over both tableaux of each shape proves the identity.
Use the <Hall inner product of symmetric functions>, with $\langle p_\lambda,p_\mu\rangle=\delta_{\lambda\mu}z_\lambda$, as defined above. The left side is the reproducing kernel $\sum_\lambda p_\lambda(x)p_\lambda(y)/z_\lambda$. Pairing its Schur expansion in the $y$ alphabet with $s_\mu(y)$ gives
$$
s_\mu(x)=\sum_\lambda s_\lambda(x)\langle s_\lambda,s_\mu\rangle.
$$
The <Schur functions> have already been proved to be a <basis>, so uniqueness of this expansion gives
$$
\boxed{\langle s_\lambda,s_\mu\rangle=\delta_{\lambda\mu}.}
$$
This is orthonormality for the Hall product; it is not a claim about an unspecified coefficientwise <inner product>.
The dual <determinant> established in Question 2 gives $\omega(s_\lambda)=s_{\lambda'}$. Apply the <symmetric-function involution> in the $y$ alphabet. Since $\omega(H_y(t))=E_y(t)$, it transforms each factor $H_y(x_i)$ into $E_y(x_i)$. The <dual Cauchy identity> is
$$
\boxed{\prod_{i,j}(1+x_iy_j)
=\sum_\lambda s_\lambda(x)s_{\lambda'}(y).}
$$
For arbitrary coefficients $f_r$ in a commutative ring, use the integral symmetric-function algebra and its free complete generators, specializing $h_r\mapsto f_r$. This works over any such coefficient ring; a literal homomorphism from the rational algebra $\Lambda$ also presupposes a $\mathbb Q$-algebra structure on the target. Specialize the first alphabet in the <Cauchy identity> to obtain the <Schur specialization by a generating function>
$$
\boxed{\prod_i F(t_i)=\sum_\lambda
\det[f_{\lambda_i-i+j}]\,s_\lambda(t)
=\sum_\lambda s_\lambda^F s_\lambda(t),}
$$
where $f_0=1$ and $f_r=0$ for $r<0$. Indeed, the left side is a product of the specialized $H(t_i)$, and the <Jacobi–Trudi identity> specializes to the displayed <determinant>. For an arbitrary infinite $F$ the result lies in the degree completion; its coefficients in each degree are well defined.
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