definitiongraph theory
I have following definitions in my Graph Theory lecture notes:
What is the difference between a trivial component and an isolated vertex? I think they are equal. Is it true?
A loop is commonly defined as an edge (or directed edge in the case of a digraph) with both ends as the same vertex. (For example from $a$ to itself). Although loops are cycles, not all cycles are loops. In fact, none of the above digraphs have any loops.
Cycles are usually defined as closed walks which do not repeat edges or vertices except for the starting and ending vertex. This definition usually allows for cycles of length one (loops) and cycles of length two (parallel edges).
Note that cycles (and walks) do not make any reference to the orientation of the edges in question. Directed cycles (and directed walks) may only travel along the "forward" direction of the edges. In particular, that implies that $G_3$ pictured above has a third cycle, $(\color{blue}{(a,b)},(b,c),(c,a))$ where the $\color{blue}{(a,b)}$ refers instead to the edge pointing from $b$ to $a$.
Technically, all of the graphs above except for $G_2$ are directed multigraphs since in each you have parallel edges. Although in simple graphs (graphs with no loops or parallel edges) all cycles will have length at least $3$, a cycle in a multigraph can be of shorter length. Usually in multigraphs, we prefer to give edges specific labels so we may refer to them without ambiguity.
As for being strongly connected, yes all of them are and your definition is correct.
Your additional question, "what is the difference between a cycle and a connected component"
The above graph contains a cycle (though not a directed cycle) yet is not strongly connected.
One can prove that if a directed multigraph is strongly connected then it contains a cycle (take a directed walk from a vertex $v$ to $u$, then a directed walk from $u$ to $v$. Any closed walk contains a cycle).
One can also show that if you have a directed cycle, it will be a part of a strongly connected component (though it will not necessarily be the whole component, nor will the entire graph necessarily be strongly connected).
When removing an edge $e=(v,w)$ of your graph there are two cases:
There exists a cycle involving $e$. In this case the cycle minus $e$ is still a path between $v$ and $w$, so any two vertices connected by a path prior to removing $e$ are still connected by a path. The number of connected components doesn't change.
There exists no cycle involving $e$, so $e$ is the only path connecting $v$ and $w$. Hence, $v$ and $w$ are now in two disjoint path components. Show that the path component containing $v$ and $w$ is split into exactly two components, while all other components stay fixed.
Having dealt with edge removal you can consider vertex removal the following way: Let $v$ be a vertex with degree $\operatorname{deg}(v)$. First remove the $\deg(v)$ edges adjacent to $v$. By the previous argument the number of connected components increases by at most $\deg(v)$. Now remove $v$ itself, which is in a connected component of its own by now, so the number of connected components decreases by $1$ when removing $v$.
Best Answer
Yes you're right. A trivial component is a connected graph with no edges so it has degree 0 and is thus an isolated vertex. However. A trivial graph is the complement of $K_n$ and that is probably what the (graph) part under 2. wants to emphasize.