Let $ER_{n} (p)$ be the Erdős-Rényi random graph (see the second model in the link), where $p=\frac{\lambda}{n}$. Furthermore, let $W_{ER_{n}}$ be the number of wedges in the graph, and $\Delta_{ER_{n}} $ be the number of triangles in the graph.
I have to show that $$\lim_{n \to \infty} n^{-1} E(W_{ER_{( \lambda / n ) }} ) = \lambda^2 \qquad (1)$$ and that $$\lim_{n \to \infty} E(\Delta_{ER_{( \lambda / n ) }} ) = \frac{1}{2} \lambda^3 \qquad(2) .$$ For (2), I computed that it's equal to $ \lim_{n \to \infty} \frac{1}{6} \big{(} \frac{\lambda}{n} \big{)}^{3} n(n-1)(n-2) = \frac{1}{6} \lambda^3$. My reasoning process went as follows: there is a probability of $\big{(} \frac{\lambda}{n} \big{)}^{3}$ that three nodes are connected, but this can be done in $6$ ways, so we divide by $6$. Furthermore, it can be done by choosing one out of $n$ nodes in the graph, which then must be connected to one of $(n-1)$ nodes in the graph, and then these two connected nodes must be connected to a third node, of which there are $(n-2)$ left.
However, apparently something went wrong with my reasoning. Can you point out where I'm wrong? And help me with deriving equation (1) ?
You are mostly right. One way to really see how/why you choose these vertices is to break up this expectation into a sum of indicator functions. For example, if $\Delta_{ER(n,\lambda/n)}$ is the number of triangles in $ER(n, \lambda/n)$, then \begin{equation*} \Delta_{ER(n, \lambda/n)} = \sum_{A=\{v_1, v_2, v_3\}} 1_{\text{triangle }v_1, v_2, v_3 \text{ is present}}, \end{equation*} where $1_{\text{triangle }v_1, v_2, v_3 \text{ is present}}$ is 1 if the triangle is present and 0 if it is not present. There are ${ n \choose 3}$ possible triangles and each one is present with probability $(\lambda/n)^3$. So by linearity of expectation, \begin{equation*} E[\Delta_{ER(n, \lambda/n)}] = \sum_{A=\{v_1, v_2, v_3\}} (\lambda/n)^3 = {n \choose 3}(\lambda/n)^3. \end{equation*}
Now as for your "wedges", which I assume is a path of length 2; note that for three vertices, there are 3 such paths. Each occurs with probability $(\lambda/n)^2$. ...