As was noted in the comments, this is a planar graph: it can be drawn in the plane without unwanted edge crossings. One elegant way was suggested by Louis: place three of the vertices at the vertices of an equilateral triangle, place the fourth vertex at the centre of the triangle, draw the sides of the triangle, and connect the central vertex to each of the corners.
There are quite a few spanning trees. Since there are four vertices, every spanning tree will have three edges (because in any tree, the number of vertices is one more than the number of edges). Those three edges must connect all four vertices, and they must not be a cycle. If you play with the graph for a while, you should see that there are two different kinds of spanning tree, snakes and claws. By a snake I mean a tree that can be stretched out into a linear graph, like this:
$\bullet----\bullet----\bullet----\bullet$
By a claw, I mean a tree in which all three edges meet at a common vertex.
Your graph has four claws, one for each of the four vertices, so the only hard part is counting the snakes. There are four snakes that use only the sides of the square: $\sqcup,\sqsubset,\sqsupset,\sqcap$. There are also snakes that use exactly two sides of the square. One looks more or less like the upper-case letter N, and another like the Cyrillic letter И; I’ll let you try to find the rest of them. Finally, there are snakes that use exactly one side of the square; one is $\ltimes$, and again I’ll let you try to find the rest.
In the last question I’m guessing that you mean the adjacency matrix of the graph. Suppose that $A=[a_{ij}]$ is the adjacency matrix of some directed graph with $n$ vertices, so that $A$ is an $n\times n$ matrix. The $i$-th row of $A$ is $a_{i1}\quad a_{i2}\quad\dots\quad a_{in}\;,$ where $a_{ij}$ is $1$ if there’s an edge from vertex $i$ to vertex $j$ and $0$ if there is no such edge. Imagine adding up these numbers one at a time. Your running total starts at $a_{i1}$; if it’s $1$, that’s because there’s an edge from vertex $i$ to vertex $1$, and if it’s $0$, that’s because there’s no such edge. Thus, at this point your running total is the number of edges leaving vertex $i$ and going to one of the first $1$ vertices. Keep going: your second running total is $a_{i1}+a_{i2}$. Can you see that this is the number of edges leaving vertex $i$ and going one of the first $2$ vertices? And that when you’re done, and your running total is the full total $a_{i1}+a_{i2}+\cdots+a_{in}$, it’s equal to the number of edges leaving vertex $i$ and going to one of the first $n$ vertices? But that’s all of the edges leaving vertex $i$, so it’s the out-degree of vertex $i$.
Now apply the same kind of reasoning to a column of $A$, say the $j$-th column; if you do it right, you’ll discover that the sum of the entries tells you something about vertex $j$.