$(a)$ Let us assume that G is a tree vertices $\geq$ 2
Trying to prove by contradiction we will assume that G does not contain atleast 2 leaves. Hence in G there are no leaves or 1 leaf.
Then, degree sum of graph $\geq$ $2(n-1) + 1$
or, $\sum_{v = 1}^{n} deg(v) \geq 2n - 1$
By handshaking lemma,
Number of edges $\geq \large \frac{2n - 1}{2}$
or,
Number of edges $\geq \large n - \frac{1}{2}$
But we know that maximum edges in a tree with n nodes can be $n-1$.
This means our assumption of of G containing less than 2 leaf nodes is incorrect.
$(b)$ Assuming G is a tree with n vertices with $n \geq 2$
We know that every vertex of a bipartite graph can be partitioned into two disjoint sets $V_{1}$ and $V_{2}$ so that every vertex of $V_{1}$ can be colored with same color and also every vertex of $V_{2}$ can be colored with another color.
Select any vertex $v$ from G. Place the vertices that are an odd distance from $v$ in $V_{1}$ and those that are an even distance from v in $V_{2}$. Clearly any two distinct vertices from $V_{1}$ are not adjacent by an edge, and likewise for $V_{1}$, because trees have no circuits; moreover, $V_{1}$ $V_{2}$ clearly partition the vertex set of the graph into two disjoint subsets. Thus, any tree is bipartite.