Dual Spaces Isomorphism 
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. 
 A: 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 $\gamma$: $[T(v)](v)\neq [\Theta_\gamma(v)](v)$.
