Ideals in ring of polynomials.
Ideals are subsets of rings and roughly work as follows: If $a$ is any element of the ideal, then $a\cdot p$ is in that ideal for every $p$ in the ring.
Therefore, (1) If $1$ is in the ideal, then for every possible $p$ in the ring, $1\cdot p$ is in the ideal. That is, every element in the ring is in the ideal, and thus the ideal is the whole ring. Conversely, if $I$ is the whole ring, then it is obvious that $1$ is in the ideal.
You can attack number (2) using the already established result (1).
Assume that there is a constant $c$ in the ideal. Then, because how this ideal works, every polynomial multiplied by $c$ is in the ideal. In particular, take the constant polynomial $c^{-1}$ (assuming $k$ is a field, this element exists). Well, this means that $c\cdot c^{-1}$ (i.e. 1) is in the ideal, and because of (1), the ideal is the whole ring. The converse is obvious: if $I$ is the whole ring, then it contains in particular every constant polynomial.
As a side comment, for this kind of proofs, if you don't know how to proceed, remember the definitions of all concepts involved (in this case, field, then ring, then ideal). If that definitions don't help you, try looking at other equivalent definitions or special cases that will help you gain intuition (for instance, the $n=1$ case above). Also, there are several techniques of solving logic implications.
If you want more generality, you could prove the following and the proof is basically the same. If $p$ is invertible in the ring (i.e., $p$ is a unit), then if $p$ belongs to an ideal $I$, then $I$ is the whole ring.
I think you have the right idea. If $1 \in I$, then $\forall f(x), 1 \cdot f(x) = f(x) \in I, \text{ so } f \in k[x_1, \ldots , x_n].$ If $c$ is a non-zero constant function with $c \in I$, then $c \cdot 1/c = 1 \in I$ so you can use the first result.