Consider the figure below:

We can find side lengths $a$ and $c$ using the Law of Sines:
$$\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R$$
where $R = 2$ is the radius of the circumscribed circle.
Since $\alpha = 45^\circ$,
\begin{align*}
a & = 2R\sin\alpha\\
& = 2 \cdot 2 \cdot \sin(45^\circ)\\
& = 4 \cdot \frac{\sqrt{2}}{2}\\
& = 2\sqrt{2}
\end{align*}
Since $\gamma = 60^\circ$,
\begin{align*}
c & = 2R\sin\gamma\\
& = 2 \cdot 2 \cdot \sin(60^\circ)\\
& = 4 \cdot \frac{\sqrt{3}}{2}\\
& = 2\sqrt{3}
\end{align*}
The area of the triangle is
$$A = \frac{1}{2}ac\sin\beta$$
since $c\sin\beta$ is the length of the altitude to side $\overline{BC}$, which has length $a$.
By the Angle Sum Theorem for Triangles,
\begin{align*}
\beta & = 180^\circ - 60^\circ - 45^\circ\\
& = 75^\circ
\end{align*}
To find the exact value of $\sin(75^\circ)$, we use the Sum of Angles Formula
$$\sin(\theta + \varphi) = \sin\theta\cos\varphi + \cos\theta\sin\varphi$$
with $\theta = 30^\circ$ and $\varphi = 45^\circ$, which yields
\begin{align*}
\sin(75^\circ) & = \sin(30^\circ)\cos(45^\circ) + \cos(30^\circ)\sin(45^\circ)\\
& = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}\\
& = \frac{\sqrt{2} + \sqrt{6}}{4}
\end{align*}
Hence, the area of triangle $ABC$ is
\begin{align*}
A & = \frac{1}{2}ac\sin\beta\\
& = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2\sqrt{3} \cdot \frac{\sqrt{2} + \sqrt{6}}{4}\\
& = \frac{1}{2} \cdot \sqrt{6}(\sqrt{2} + \sqrt{6})\\
& = \frac{1}{2}(\sqrt{12} + 6)\\
& = \frac{1}{2}(2\sqrt{3} + 6)\\
& = \sqrt{3} + 3
\end{align*}
square units.