Solution: We recall that if $x = a$ is a root of $f(x)$ with multiplicity $m$, then \[ f(x) = (x-a)^m g(x),\quad g(a)\neq 0. \] This implies $f(a) = 0$ and $f^{(i)}(a) = 0$ for $1\leq i\leq m-1$ and $f^{(m)}(a) \neq 0$. Here we have $f(1) = 0 = f(3)$. Consider \begin{align*} f'(x) & = 5x^4 + 16x^3 + 12x^2 + 8x + 1 \\ & \equiv \left(x^3 + 2x^2 + 3x + 1\right) \ (\mathrm{mod}\ 5) \\ \implies f'(1) & = (1 + 2 + 3 + 1) \equiv 2\ (\mathrm{mod}\ 5) \\ \implies f'(3) & = (27 + 18 + 9 + 1) \equiv 0 (\mathrm{mod}\ 5). \end{align*} Thus, the multiplicity of $x= 1$ is $1$. Similarly, \begin{align*} f''(x) & = 3x^2 + 4x + 3 \\ \implies f''(3) & = 27 + 12 + 3 \equiv 42 (\mathrm{mod}\ 5) \equiv 2 (\mathrm{mod}\ 5). \end{align*} Thus, the multiplicity of $5$ is $2$. Therefore the correct answer is Option D.