Past exam of the mathematical methods I course of the University of Oxford / 2015 resit / 7

Yes. The two columns form the matrix $\left(\begin{array}{cc}4& 1\\ 0& -1\end{array}\right)$, whose determinant is $-4\ne 0$. They are therefore linearly independent and span the two-dimensional real vector space, so they form a basis.

No, as printed. Each displayed vector has only two components, so neither is an element of the stated ${\mathbb{R}}^3$ without an additional embedding convention. Even if one inserts a zero third component, two vectors span at most a two-dimensional linear subspace and cannot form a basis of ${\mathbb{R}}^3$. This conclusion does not require guessing a missing third entry.

No. Calling the three vectors $v_1,v_2,v_3$ in order, direct addition gives $v_3=v_1+v_2$. This nontrivial linear dependence prevents them from forming a basis of ${\mathbb{R}}^3$. The first two are not proportional, so the span has dimension two.

Yes, over $\mathbb{C}$. The column matrix is $\mathrm{diag}(1,1+i)$, with nonzero determinant $1+i$. Every $(z_1,z_2)\in {\mathbb{C}}^2$ has the unique expansionThe scalar field matters: the same space regarded as a real vector space has dimension four.