I'll answer question 1 only for now, but I might edit this to address the others later.
One should note that corresponding to any set of orthogonal polynomials, there exists a symmetric tridiagonal matrix, called a Jacobi matrix, whose characteristic polynomial is the monic (leading coefficient is 1) version of the set of orthogonal polynomials considered. To use the Legendre polynomials as an explicit example, we first note that the monic Legendre polynomials satisfy the following two-term recurrence relation:
$$\hat{P}_{n+1}(x)=x \hat{P}_n(x)-\frac{n^2}{4 n^2-1}\hat{P}_{n-1}(x)$$
where $\hat{P}_n(x)=\frac{(n!)^2 2^n}{(2n)!}P_n(x)$ is the monic Legendre polynomial.
From this, we can derive an explicit expression for the corresponding Jacobi matrix (here I give the 5-by-5 case):
$$\begin{pmatrix}0&\frac{1}{\sqrt{3}}&0&0&0\\\frac{1}{\sqrt{3}}&0&\frac{2}{\sqrt{15}}&0&0\\0&\frac{2}{\sqrt{15}}&0&\frac{3}{\sqrt{35}}&0\\0&0&\frac{3}{\sqrt{35}}&0&\frac{4}{\sqrt{63}}\\0&0&0&\frac{4}{\sqrt{63}}&0\end{pmatrix}$$
(the general pattern is that you have $\frac{n}{\sqrt{4 n^2-1}}$ in the $(n,n+1)$ and $(n+1,n)$ positions, and 0 elsewhere.)
We now note that $\frac{n}{\sqrt{4 n^2-1}}$ can never be 0, and then use the fact that if a symmetric tridiagonal matrix has no zeroes in its sub- or superdiagonal, then all its eigenvalues have multiplicity 1. (A proof of this fact can be found in Beresford Parlett's The Symmetric Eigenvalue Problem.) Thus, all the roots of the Legendre polynomial are simple roots.
A more conventional proof of this fact is in page 27 of Theodore Chihara's An Introduction to Orthogonal Polynomials. Briefly, the argument is that $P_n(x)$ changes sign at least once within $[-1,1]$ (and thus has at least one zero of odd multiplicity within the support interval) since
$$\int_{-1}^1 P_n(u)\mathrm du=0$$
Now, the polynomial
$$P_n(x)\prod_{j=1}^k(x-\xi_j)$$
where the $\xi_j$ are the distinct zeroes of odd multiplicity within $[-1,1]$, should be greater than or equal to zero within $[-1,1]$, and thus its integral over $[-1,1]$ should be greater than zero. However, since
$$\int_{-1}^1 P_n(u) u^k\mathrm du=0\qquad\text{if}\qquad k < n$$
we have a contradiction, and thus all the roots of the Legendre polynomial are simple (and within the support interval $[-1,1]$).