= Solution
For the finite <probability distribution>, put $\xi_j=(X_j+1)/2$. These are <independent> fair zero-one variables, and
$$
Y_n=2\sum_{j=1}^n2^{-j}\xi_j-(1-2^{-n}).
$$
The binary integer $\sum_{j=1}^n2^{n-j}\xi_j$ is uniform on $\{0,\ldots,2^n-1\}$. Hence
$$
\boxed{\mathbb P\!\left(Y_n=-1+\frac{2k+1}{2^n}\right)=2^{-n},\qquad 0\leq k<2^n.}
$$
These are the midpoints of $2^n$ equal subintervals of $[-1,1]$. For every bounded continuous $f$, the <Riemann sums> give
$$
\mathbb E f(Y_n)=2^{-n}\sum_{k=0}^{2^n-1}f\!\left(-1+\frac{2k+1}{2^n}\right)\longrightarrow\frac12\int_{-1}^1f(x)\,dx.
$$
Thus $\boxed{Y_n\Rightarrow\operatorname{Unif}[-1,1]}$ in the sense of <weak convergence of probability measures>.
For the almost-sure and $L^2$ assertions, the <Dyadic Rademacher series> $Y=\sum_{j\geq1}2^{-j}X_j$ converges absolutely on every sign sequence, since $|X_j|=1$. Its tail satisfies $|Y-Y_n|\leq2^{-n}$, proving <almost sure convergence> and <convergence in L2>. More precisely, zero means and <independence> eliminate the cross terms:
$$
\boxed{\mathbb E[(Y-Y_n)^2]=\sum_{j>n}4^{-j}=\frac{4^{-n}}3.}
$$
This identity follows first for finite tails and then by <dominated convergence>. Bounded continuous test functions and <bounded convergence theorem> identify the <probability distribution> of $Y$ with the weak limit above, namely the <uniform distribution> on $[-1,1]$.
For the requested product identity, each <Rademacher random variable> has <characteristic function> $\mathbb E e^{itX_j}=\cos t$. The <characteristic function of a sum of independent variables> therefore yields
$$
\varphi_{Y_n}(t)=\prod_{j=1}^n\cos(t/2^j).
$$
By part (a), or directly by <bounded convergence theorem>, this converges to the <characteristic function of a uniform distribution>, $\tfrac12\int_{-1}^1e^{itx}\,dx=\sin t/t$. Consequently the <dyadic cosine product> is
$$
\boxed{\prod_{j=1}^{\infty}\cos(t/2^j)=\frac{\sin t}{t},\qquad t\in\mathbb R.}
$$
Both sides are one at zero. At zeros of the <sinc function> the identity is understood as the limit of the finite products, so no logarithm or division by a vanishing factor is needed.
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