On the definition of a reductive group. Wikipedia1 defines a reductive group $G$ as an algebraic group with trivial unipotent radical. The radical is the connected component of identity in the maximal normal solvable subgroup of $G$. The unipotent radical is the set of the unipotent elements of the radical. An element $u$ is unipotent if $(u-1)$ is nilpotent.
Question 1 - This last line is where everything breaks down in my head. If we talk of unipotent then there is a clear difference between what $1$ is and what $0$ is. In this case we have to be in a ring. But here we are just in a group $(G)$. What am I missing?

EDIT -
Question 2 - In this book (chapter 5) I have come across the definition that an affine algebraic group $G$ is reductive if its maximal connected solvable subgroup is a torus. (This is what Tobias mentions in his comment). But I am not sure how the definitions are equivalent when $G$ is affine.
Thank you.
1Link to the revision of the Wikipedia article at the time of this post.
 A: I don't know how to answer for algebraic groups in general, only affine algebraic groups over an algebraically closed field $k$.  Remember that an affine algebraic group is up to isomorphism the same thing as a closed subgroup of $\textrm{GL}_n$ for some $n$.  However, an isomorphism of such a group into $\textrm{GL}_n$ is not canonical.
Question 1: For any matrix $x \in \textrm{GL}_n$, there exist unique $x_s, x_u \in \textrm{GL}_n$ such that $x = x_sx_u = x_ux_s$, $x_s$ is diagonalizable, and $x_u$ is unipotent.  We call $x_u, x_s$ the unipotent and semisimple components of $x$, respectively.  This is the multiplicative version of Jordan canonical form.
Let $G$ be such a LAG over $k$.  If $x \in G$, there exist unique $x_s, x_u \in G$ such that for any isomorphism $\varphi$ of $G$ onto a closed subgroup of $\textrm{GL}_n$ for some $n$, $\varphi(x_u), \varphi(x_s)$ are the unipotent and semisimple components of $\varphi(x)$.  This is proved in Chapter 2 of Springer, Linear Algebraic Groups.  Thus an $x \in G$ being unipotent doesn't depend on how you regard $G$ as a closed subgroup of $\textrm{GL}_n$.
Let $R(G)$ be the radical of $G$: it is the unique maximal closed, connected, solvable, normal subgroup of $G$ (the notes you cited must have a typo; e.g. $\textrm{GL}_n$ is reductive, the upper triangular invertible matrices are a maximal connected closed solvable group, but that's not a torus).
Question 2: $R_u(G)$ is trivial if and only if $R(G)$ is a torus.
If $R(G)$ is a torus, then $R_u(G)$ (which is equal to the set of unipotent elements $R(G)_u$ of $R(G)$) is trivial, because it is a unipotent subgroup of a semisimple group.  
Suppose $R_u(G) = R(G)_u$ is trivial.  Now $R(G)$ is a connected solvable group.  If $H$ is a connected solvable group, and $T$ is a maximal torus of $H$, then as a variety, $H$ is isomorphic to the product of $T$ and the set $H_u$ of unipotent elements of $H$ (6.3, Springer).  If $H_u$ is trivial, that says that $T = H$. So a connected solvable group with no nontrivial unipotent elements is automatically a torus.
