Let $G = (V, E)$ be a finite graph and let $\Delta$ denote its Laplacian (here we take the convention that the Laplacian is positive semidefinite). For simplicity name the vertices $1, 2, ... n$. Then $\Delta$ is the matrix associated to the quadratic form
$$q(x_1, ... x_n) = \sum_{(i, j) \in E} (x_i - x_j)^2$$
which one might call the Dirichlet energy. Recall that, for an appropriate change of variables, such a quadratic form can be written as
$$q(z_1, ... z_n) = \sum_i \lambda_i z_i^2$$
where $\lambda_i$ are the eigenvalues of $\Delta$. Thus to determine the multiplicity of the zero eigenvalue it suffices to determine when $q$ can be zero. But from the first expression it should be clear that $q = 0$ if and only if $x_i = x_j$ whenever $(i, j) \in E$; that is, whenever $(x_1, ... x_n)$ determines a function
$$x : V \to \mathbb{R}$$
which is constant on each connected component of $G$. The dimension of this space of functions is clearly the number of connected components of $G$, and the conclusion follows.
If you put all 1 on the diagonal of your adjacency matrix $A$, and all edge weights are positive then when you multiply $A^2 = A*A$ you get a non-zero entry $a_{ij}$ in $A^2$ if and only if there exist non-zero $a_{ik}$ and $a_{kj}$ in $A$ for some $k$, i.e. there is a path of length $2$ between $i$ and $j$ if $k\neq j$ and $k\neq i$ and there is a path of length $1$ if $k = j$ or $k = i$. So the non-zero entries in $A^2$ tell you all pairs of nodes that are connected by a path of length $2$. Similarly the entries in $A^k$ tell you all pairs of nodes that are connected by a path of length $k$. So if you start with $A$ and keep squaring until you get $A^k$ where $k \geq n$ where $n$ is the number of nodes, then the non-zero entries in row $i$ tell you all the nodes that are connected to node $i$ (since two connected nodes must be connected by a path of length $n$ or less). So if you have a row in $A^k$ that is all non-zero, then the graph is connected. If the graph is not connected, you can similarly tell the connected components from the rows of $A^k$.
Best Answer
If the distinct eigenvalues of the adjacency matrix of a $k$-regular graph are $k=\theta_1\ge\cdots\ge\theta_m$, the eigenvalues of its Laplacian in non-decreasing order are \[ 0, k-\theta_2,\ldots,k-\theta_m. \] So in this case the algebraic connectivity and the spectral gap coincide.
If the graph is not regular then, in general, there is no simple or useful relation between the eigenvalue of the adjacency matrix and the eigenvalues of the Laplacian.