iii) It won't be isometric, but its image will be closed.
This stems from the following result:
Let $A$ and $B$ be $C^*$-algebras, and let $\varphi:A\to B$ be a $*$-homomorphism. If $\varphi$ is injective, then it is an isometry.
A proof of this can be found in most introductory books on $C^*$-algebras, for example, in chapter 1 of Davidson's $C^*$-Algebras by Example, or in chapter 3 of Murphy's $C^*$-Algebras and Operator Theory. This gives i).
The positive answer for ii) follows from the positive answer to i) and from a standard result in functional analysis:
If $X$ and $Y$ are Banach spaces and $T:X\to Y$ is an isometric linear map, then $T(X)$ is closed.
To answer iii), we use a corollary of the first result:
Let $A$ and $B$ be $C^*$-algebras, and let $\varphi:A\to B$ be a (not necessarily injective) $*$-homomorphism. Then $\varphi(A)$ is a $C^*$-subalgebra of $B$ (i.e., its image is closed).
This comes from the fact that the image of $\varphi:A\to B$ is the same as the image of the induced $*$-homomorphism $\tilde\varphi:A/\ker(\varphi)\to B$, which is injective, hence isometric, hence has closed range.