Help with proof involving weighted full binary tree
There are two basic induction patterns for (non-empty) full binary trees:


A tree is either a leaf or consists of a root and two full binary subtrees.
A tree is either a leaf or can be obtained from a smaller full binary tree by adding two children to a leaf.


Your first question can be proved in both ways. In both cases, the base case is clear: if a tree is a leaf then it has $n=1$ leaves and $n-1=0$ internal nodes.

According to the first way, if the tree $T$ is not a leaf then it has a root and two subtree $T_1,T_2$. By induction we know that if $T_1,T_2$ have $n_1,n_2$ leaves (respectively) then they have $n_1-1,n_2-1$ internal nodes (respectively), and together with the root, this gives $n_1+n_2-1$ internal nodes for $T$, which is exactly one less than the total number of leaves.

According to the second way, the tree $T$ is obtained from a smaller tree $T'$ by adding two children to a leaf. If $T'$ has $n'$ leaves and $n'-1$ internal nodes, then $T$ has $n'-1+2 = n'+1$ leaves and $n'-1+1 = n'$ internal nodes, exactly one less than the number of leaves.

Your other statement can also be proved in both ways. I leave you the details.