Hello Oinker,
$(a)\;$ Eigenvalues of $A=\begin{bmatrix}{2}&{3}\\{4}&{1}\end{bmatrix}$: $$\chi_A(\lambda)=\lambda^2-(\mbox{trace }A)\;\lambda+\det A=\lambda^2-3\lambda-10=0\\\Leftrightarrow \lambda=5\mbox{ (simple) }\vee\;\lambda=-2\mbox{ (simple) }$$ Eigenvectors and corresponding basis: $$\ker(A-5I)\equiv\left \{ \begin{matrix} -3x_1+3x_2=0\\4x_1-4x_2=0\end{matrix}\right.,\qquad B_5=\{(1,1)^T\}$$ $$\ker(A+2I)\equiv\left \{ \begin{matrix} 4x_1+3x_2=0\\4x_1+3x_2=0\end{matrix}\right.,\qquad B_{-2}=\{(3,-4)^T\}$$
$(b)\;$ If $\lambda=0$ is eigenvalue of $B$, then $\det (B-0I)=0$ so, $\det B=0$. As a consequence, $B$ is not invertible.
$(c)\;$ If $B$ is not invertible, $0=\det (B)=\det(B-0I)$ which implies that $\lambda=0$ is an eigenvalue of $B$.
$(d)\;$ According to $(b)$ and $(c)$, if $\lambda$ is an eigenvalue of $B$, necessarily $\lambda\neq 0$ and there exists $0\ne x\in K^n$ suach that $Bx=\lambda x$. Then, $$Bx=\lambda x\Rightarrow B^{-1}(Bx)=B^{-1}(\lambda x)\Rightarrow Ix=\lambda (B^{-1}x)\Rightarrow B^{-1}x=\frac{1}{\lambda}x$$ which implies that $1/\lambda$ is an eigenvalue of $B^{-1}$.