By Stacks, Lemma 063U, if $f: X \rightarrow Y$ is a smooth surjective morphism, with $X$ a smooth variety over a field $k$ (seen as a scheme) and $f$ is a $k$-morphism, then $Y$ is smooth over $k$ at any point in the image of $f$.
This suggests the following counterexample for 2: let $X =\mathbb{A}^1\backslash \{0\}$ (with coordinate $t$) and $Y=\operatorname{Spec}\,k[x,y]/(x^2-y^3)$, and $X \rightarrow Y$ given by $(x,y) \longmapsto (t^3,t^2)$.
$Y$ is smooth at every point except the maximal ideal $(x,y)$, that is, $Y$ is smooth at exactly the points in the image of the morphism $X \rightarrow Y$.
Now, consider a map $f: X \rightarrow Y$ of smooth schemes over a field $k$. Let $x \in X$ be a point. Then the following are equivalent:
- $f$ is smooth at $x$
- the natural map $\Omega^1_{Y/k,f(x)} \_{O_{Y,f(x)}} O_{X,f(x)} \rightarrow \Omega^1_{X/k,x}$ (where $\Omega^1$ denotes the sheaf of relative differentials) is an isomorphism onto a direct factor.
- $\Omega^1_{Y/k} \otimes_{O_{Y,f(x)}} \kappa_X(x)\rightarrow \Omega^1_{X/k,x} \otimes_{O_{X,f(x)}} \kappa_X(x)$ is injective, where $\kappa$ denotes the residue field at a point.
- The tangent map to $f$ at $x$ is surjective.
(See Neron models by Bosch, Lutkebohmert and Raynaud, Chapter 2.3, Prop 8 – I added point 4, it is pretty elementary to derive and seems the most intuitive).