So for the operator $A:l_2(\Bbb C)\to l_2(\Bbb C)$ defined as:
$$A(x_1,x_2,\cdots,x_m,x_{m+1},x_{m+2},\cdots) = (x_1,x_2,\cdots,x_m,0,0,\cdots)$$
We can find the adjoint operator $A^*$ by looking at:
$$\langle Ax,y\rangle = \sum_{i=1}^m x_i\overline{y}_i = \sum_{i=1}^\infty x_i A^*(\overline{y})_i$$
$\implies A^*(\overline{y})_i = (\overline y_1, \overline y_2,\cdots,\overline y_m,0,\cdots)$
So then: $$\langle Ax,y\rangle = \sum_{i=1}^m x_i\overline{y}_i = \sum_{i=1}^\infty x_i A^*(\overline{y})_i = \sum_{i=1}^m x_i\overline y_i=\langle x,A^*y\rangle$$
Is that correct?
Now how do I find the spectrum of $A$ and the resolvent of $A$?
The operator $A$ is the orthgonal projection onto the span of $\{ e_1,e_2,\cdots,e_m\}$, where $e_j$ is the j-th standard basis element of $\ell^2$. So $A^2=A=A^{\star}$. To invert $A-\lambda I$, the inversion can be done separately on the range of $A$ and on the null space of $A$ because these are invariant under $A$ and mutually orthogonal, with $\ell^2=\mathcal{N}(A)\oplus\mathcal{R}(A)$. On $\mathcal{N}(A)$, $(A-\lambda I)$ is $-\lambda I$. On $\mathcal{R}(A)$, $(A-\lambda I)$ is $(1-\lambda)I$. So, the inverse is $$ (A-\lambda I)^{-1} = -\frac{1}{\lambda}(I-A)+\frac{1}{1-\lambda}A. $$ You can directly verify that the above holds for $\lambda\notin\{0,1\}$; this requires knowing only that $A^2=A$. That is, you can verify that $$ \left[-\frac{1}{\lambda}(I-A)+\frac{1}{1-\lambda}A\right](A-\lambda I)=I,\\ (A-\lambda I)\left[-\frac{1}{\lambda}(I-A)+\frac{1}{1-\lambda}A\right]=I. $$ And you can verify that $A-\lambda I$ is not invertible for $\lambda=0$ or $\lambda=1$ because $\mathcal{N}(A)\ne \{0\}$ and $\mathcal{N}(A-I)\ne \{0\}$. So the spectrum of $A$ is $\sigma(A)=\{0,1\}$. This is true of any non-trivial projection.