So I have to answer the above question as part of homework. I am completely confused by the idea of dual spaces; here, where does the (x,y) term come in? I don't even know how the isomorphism exists here; how do I go about forming a new basis for part (a)? Any help would be much appreciated.
As Ivo Terek observed, to define $\Theta_\alpha(a,b)$ you have to know how it acts on a vector of $\mathbb R^2$, because it is a linear functional.
(a): First I give you a base that works; second I try to characterize all the basis that works (in order to explain how I have found the first one).
Take the base $\beta = \{(-1,0), (0,1)\}$. I claim that $\Theta_\beta = \Theta_\alpha$. For, we have to show $[\Theta_\beta(a,b)](x,y) = ax+by$. Note that: \begin{gather} (a,b) = -a(-1,0) + b(1,0),\qquad (x,y)=-x(-1,0) + y(1,0) \end{gather} So now we can compute the value of $[\Theta_\beta(a,b)](x,y)$: \begin{gather} [\Theta_\beta(a,b)](x,y) = [\Theta_\beta(-a(-1,0) + b(1,0))](x,y)=\\ [-af_1+bf_2](x,y)=\\ -af_1(x,y) + bf_2(x,y)=\\ -af_1(-x(-1,0) + y(1,0)) + bf_2(-x(-1,0) + y(1,0)) = \\ (-a)(-x) + by = ax+by \end{gather} Where I used the linearity of $\Theta_\beta$ and the definition and linearity of $f_1,f_2$. So the base $\beta$ works.
Now I'll show how I found this base. Take a generic base $\beta=\{x_1,x_2\}$. Since $\beta$ is a base there exist $m,n,p,q\in \mathbb R$ such that: $$ (a,b) = mx_1 + nx_2,\qquad (x,y)=px_1 + qx_2 $$ Now compute the value of $[\Theta_\beta(a,b)](x,y)$ in the same as above founding: \begin{equation} [\Theta_\beta(a,b)](x,y) = mp +nq \end{equation} Since we want $\Theta_\alpha = \Theta_\beta$ this is possibile if and only if: \begin{gather} ax+by = mp+nq\\ \langle (a,b)^T, (x,y)^T\rangle =\langle (m,n)^T, (p,q)^T\rangle \end{gather} where $\langle \cdot, \cdot\rangle$ denotes the standard scalar product of $\mathbb R^2$. If we call $A = (x_1^T | x_2^T)$ the matrix of change of coordinates between the base $\beta$ and $\alpha$ we have: $$ (a,b)^T = A(m,n)^T, \qquad (x,y)^T = A(p,q)^T $$ Hence the condition $[\Theta_\beta (a,b)](x,y) = [\Theta_\alpha (a,b)](x,y)$ is equivalent to require that the matrix $A$ is orthogonal!
Hence you can choose for $\beta$ any orthonormal base, and you obtain the statement (and this is the reason I choosed $\beta = \{(-1,0),(0,1)\}$ at the beginning.
(b): As suggested again by Ivo Terek, let's compute $[T(v)](v)$ and $[\Theta_\gamma(v)](v)$ for a generic base $\gamma$ and $v\neq 0$: \begin{gather} [T(v)](v) = [T(a,b)](a,b) = ab-ab = 0\\ [\Theta_\gamma(v)](v)= [\Theta_\gamma(a,b)](a,b) = m^2+n^2 \end{gather} Where the last computation is obtained observing what we did in the part (a). Since $v\neq 0$, then $m^2+n^2>0$. Hence $T$ does not coincide with $\Theta_\gamma$ for any base $\gamma$: $[T(v)](v)\neq [\Theta_\gamma(v)](v)$.