How to find the multiplicity of eigenvalues?

The characteristic polynomial of the matrix is $p_A(x) = \det (xI-A)$. In your case, $A = \begin{bmatrix} 1 & 4 \\ 2 & 3\end{bmatrix}$, so $p_A(x) = (x+1)(x-5)$. Hence it has two distinct eigenvalues and each occurs only once, so the algebraic multiplicity of both is one.

If $B=\begin{bmatrix} 5 & 0 \\ 0 & 5\end{bmatrix}$, then $p_B(x) = (x-5)^2$, hence the eigenvalue $5$ has algebraic multiplicity $2$. Since $\dim \ker (5I-B) = 2$, the geometric multiplicity is also $2$.

If $C=\begin{bmatrix} 5 &1 \\ 0 & 5\end{bmatrix}$, then $p_C(x) = (x-5)^2$ (same as $p_C$), hence the eigenvalue $5$ has algebraic multiplicity $2$. However, $\dim \ker (5I-C) = 1$, the geometric multiplicity is $1$.

Very loosely speaking, the matrix is 'deficient' in some sense when the two multiplicities do not match.

The algebraic multiplicity of an eigenvalue $\lambda$ is the power $m$ of the term $(x-\lambda)^m$ in the characteristic polynomial.

The geometric multiplicity is the number of linearly independent eigenvectors you can find for an eigenvalue.

Let me explain the two multiplicities that I know are related to eigen-values of matrices:
Firstly, what is the eigenvalue of a matrix $A$? By definition it consists of the zeros of the polynomial: $\det(A-xI)$. So the muliplicities that they occur in this polynomial are defined to be the multiplicities of the eigen-values.
Secondly, since, for an eigen-value $\lambda$, we have $\det(A-\lambda I)=0$, i.e. $A-\lambda I$ is a singular matrix, and the linear transformation it defines has a non-trivial kernel. The dimension of this kernel is then said to be the geometric multiplicity of the eigen-value.
Hence, in one case, one has to compute some polynomial; while, on the other hand, one has to compute some transformations, to find its kernel, and to determine the dimension of the kernel, to find the multiplicites of eigen-values. Notice here that you have a $2\times 2$ matrix with $2$ eigen-values, and hence they must be of multiplicities $1$. That is to say, the two notions coincide, and the matrix in question must be diagonalisable.