Eigenvalues of Symmetric Tridiagonal Matrices
The type of matrix you have written down is called Jacobi matrix and people are still discovering new things about them basically their properties fill entire bookcases at a mathematics library. One of the reasons is the connection to orthogonal polynomials. Basically, if $\{p_n(x)\}_{n\geq 0}$ is a family of orthogonal polynomials, then they obey a recursion relation of the form $$ b_n p_{n+1}(x) + (a_n- x) p_n(x) + b_{n-1} p_{n-1}(x) = 0. $$ You should be able to recognize the form of your matrix from this.
As far as general properties of the eigenvalues, let me mention two:
The eigenvalues are simple. In fact one has $\lambda_j - \lambda_{j-1} \geq e^{-c n}$, where $c$ is some constant that depends on the $b_j$.
The eigenvalues of $A$ and $A_{n-1}$ interlace.
Amongst the polynomials that can arise as characteristic polynomials of tridiagonal matrices with zero diagonal, one finds the Hermite polynomials. Schur showed that Hermite polynomials of even degree are irreducible and that their Galois groups are not solvable. Hence there can be no closed form expression for the zeros in terms of the $b_i$'s in general.
According to doi:10.1016/S0024-3795(99)00114-7, the closed form for the eigenvalues of a tridiagonal Toepliz matrix of the form
$$ \begin{bmatrix}a & b\\ c & a & b\\ & \ddots & a & \ddots \\ & & & \ddots & \\ & & & c & a \end{bmatrix} $$
is:
$$\lambda_{k}=a+2\sqrt{bc}\cos\left[\frac{k\pi}{(n+1)}\right], \quad k=1\cdots n $$