= Solution
We first give an auxiliary-integral proof, rather than deducing both requested results from a theorem only stated later. The following <symmetric Hermite integral obstruction> will do both jobs. Suppose
$$
k+\sum_{j=1}^s b_je^{\beta_j}=0,
$$
where $k\ne0$ and $b_j$ are <integers>, the distinct nonzero algebraic exponents form a Galois-stable set, and their weights are invariant under conjugation. Choose $P\in\mathbb Z[X]$ vanishing at those exponents, with $P(0)\ne0$, and a positive <integer> $c$ for which every $c\beta_j$ is an <algebraic integer>. For a large <prime number> $p$, let
$$
D=(\deg P+1)p-1,\qquad f_p(X)=\frac{c^DX^{p-1}P(X)^p}{(p-1)!},\qquad F_p(X)=\sum_{r=0}^D f_p^{(r)}(X).
$$
At zero, derivatives below $p-1$ vanish; the derivative of order $p-1$ equals $c^DP(0)^p$. All higher derivatives at zero are <integers> divisible by $p$. At each $\beta_j$, derivatives below $p$ vanish and all higher derivatives belong to $p\mathcal O_L$ in a Galois <splitting field> $L$. Indeed their coefficients contain $r!/(p-1)!$, divisible by $p$ for $r\ge p$, and every $c^D\beta_j^u$ is integral for $u\le D$. Consequently
$$
I_p=kF_p(0)+\sum_jb_jF_p(\beta_j)\in\mathbb Z,\qquad I_p\equiv kc^DP(0)^p\pmod p.
$$
Galois invariance makes the sum rational, and integrality makes it an <integer>. For <prime numbers> not dividing $kcP(0)$ it is nonzero.
But $F_p-F_p'=f_p$, so integration along the straight segment to $\beta$ gives
$$
e^\beta F_p(0)-F_p(\beta)=\int_0^\beta e^{\beta-z}f_p(z)\,dz.
$$
Multiply by the weights and use the assumed exponential relation. This expresses $-I_p$ as the sum of these integrals. On the finitely many fixed segments, their absolute values are at most $C^p/(p-1)!$ for a fixed $C$: all <polynomial> and exponential factors have fixed bounds, and $D$ is linear in $p$. Thus $|I_p|\to0$, contradicting its being a nonzero <integer>. The obstruction is proved.
If $e$ were algebraic, an integral <polynomial> relation would give $a_0+\sum_{j=1}^s a_je^j=0$ with $a_0\ne0$. <Integer> exponents and weights satisfy the obstruction, so \b[$e$ is transcendental].
If $\pi$ were algebraic, put $\alpha=i\pi$ and let $\alpha_1,\ldots,\alpha_d$ be its algebraic conjugates. Since $1+e^\alpha=0$, the product $\prod_j(1+e^{\alpha_j})$ vanishes. Expand it and collect equal subset sums. The zero sums contribute a positive <integer> $k\ge1$; the nonzero sums give a Galois-stable list $\beta_j$ with positive integral multiplicities. This is precisely the prohibited relation. Hence \b[$\pi$ is transcendental].
The general <Lindemann–Weierstrass theorem> states that exponentials of distinct <algebraic numbers> are linearly independent over $\overline{\mathbb Q}$. Suppose the sine quotient were algebraic, say $\gamma$. The denominator is nonzero: $\sin\beta=0$ would give a forbidden relation between the exponentials of the distinct <algebraic numbers> $i\beta,-i\beta$. The quotient identity would give
$$
e^{i\alpha}-e^{-i\alpha}-\gamma e^{i\beta}+\gamma e^{-i\beta}=0.
$$
The four exponents are distinct under the stated nonzero and non-opposite assumptions, so this also contradicts <linear independence>. Therefore
$$
\boxed{\sin\alpha/\sin\beta\text{ is transcendental}}.
$$
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