Graphs with a given number of edges and vertices of degree $3$ Original problem:

Let a $\Gamma$ be a set of all graph's G following rules: $V_G \subset \mathbb{N}$, $|E_G| = 303$ and $G$ has $151$ vertices of degree $3$. (Last condition means that G can have more vertices of other degrees.) Decide and argument which of these following statements are true:
a) Graph $G\in \Gamma $ is connected if and only if $G$ is a tree.
b) If $G\in\Gamma$ has a vertex of a degree $7$, then $G$ is not a tree.
c) If $G\in\Gamma$ is  disconnected and does not have any isolated vertices, then $G$ has a cycle.

My thoughts:

a) I know that a tree is connected. I also know that for a graph to be connected it has to have an edge between each neighboring vertices. So I think for that answer it might be a no, because some other graphs could possibly be connected? 
b) A tree has $|E_G| + 1 = |V_G|$, so it must have $|V_G| = 304$. And then it is not possible to construct a tree following the rules. Because every two vertices in a tree are connected exactly one path. So to have a tree with $151$ vertices of a degree $3$ we would need at least $151 * 3 + 1$ edges, and thus it would not satisfy condition of $|E_G| = 303$.

c) So to have a cycle we need all vertices in that cycle of a degree 2. But I'm not really sure on how to go further.

EDIT:
a) Is not true.
Here is how a graph with 151 vertices of degree 3, 2 vertices of degree 2 (lower right sub-graph) and 269 edges looks like:
graph with 151 vertices of degree 3
so we need to add 303 - 269 edges to met all conditions which we can do in following way:
full graph with met conditions
We have showed that the graph is connected and is not a tree
b) Is true.

*

*A tree has $|E_G| + 1 = |V_G|$, so it must have $|V_G| = 304$.

*A sum of all degrees in a tree is $2 * |E_G| = 606 $

*A tree can have at most $\lfloor \frac{n-2}{k-1}\rfloor - 1$ degrees of a degree k and n number of vertices.
In our case it is $\lfloor \frac{304}{2} - 1 \rfloor = 151$
If a tree has 151 vertices of a degree 3, then sum of these degrees is 151 * 3 = 453, so we are left with 606 - 453 = 153 degrees among 304 - 151 = 153 vertices. Which is not possible to construct a tree with a vertex of degree 7 anymore.
c) Still not sure how to approach.
Are my thoughts correct? If not, what am I doing wrong?
 A: I would suggest this counterexample to problem (a). Let $C_{151}$ be a cycle of order $151$ and $u_i$ be its vertices. Let two additional vertices $v$ and $w$ other than $u_i$ be chosen.  Now construct graph $G$. Its vertices $V(G)=\{u_i,v,w\mid i=1,\ldots,151\}$. Its edges are edges of the cycle $C_{151}$ and still $u_iv$, $i=1,\ldots,151$,  $vw$.
Consequently, graph $G$ has $303$ edges, $151$ vertices of degree $3$, is obviously connected, and not a tree, since it has a cycle.
The solution (b) is generally correct, but it is incomplete.
So, if $G$ is a tree, then besides $151$ vertices of degree $3$, there must be $153$ other vertices of degree $\geq1$.
Let's denote these additional vertices by $x_i$, $i=1,\ldots,153$. Since $\sum_{i=1}^{153}\operatorname{deg}(x_i)=153$ and $\operatorname{deg}(x_i)\geq1$, it follows
that $\operatorname{deg}(x_i)=1$ for all $i$. So if $G$ is a tree, then it has vertices of degree $3$ and $1$, but not $7$.
Now, it is easy to answer question (c).
Indeed, suppose $G$ is disconnected and does not have any isolated vertices and has no cycles. Then each component of graph $G$ is a tree.
Denote these trees by the symbol $T_i$, $i=1,\ldots,k$, and $k\geq2$.
Let still $t_i$ be the number of vertices of degree $3$ in $T_i$.
By assumption we have:
\begin{align*}
&\sum_{i=1}^k t_i=151,\\
&\sum_{i=1}^k|E(T_i)|=303,\\
&\sum_{i=1}^k|V(T_i)|=\sum_{i=1}^k(|E(T_i)|+1)=303+k,\\
&\sum_{i=1}^k(|V(T_i)|-t_i)=303+k-151=152+k>153,\\
&606=\sum_{i=1}^k\sum_{x\in T_i}\operatorname{deg}(x)=\sum_{i=1}^k3t_i+\sum_{i=1}^ks_i=453+\sum_{i=1}^ks_i,\\
&\Rightarrow\ \sum_{i=1}^ks_i=153.
\end{align*}
where $s_i$ is the sum of degrees of those vertices $T_i$ whose degree is different from three.
So we have at least $154$ vertices whose sum of powers is $153$. Contradiction.
