Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.

Given: Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.

Diagonals of a parallelogram bisect each other.

∴ Mid point of AC = Mid point of BD

Mid point of $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is $\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$.

Mid point of $A C=\left(\frac{3+a}{2}, \frac{1+b}{2}\right)$

Mid point of $B D=\left(\frac{5+4}{2}, \frac{1+3}{2}\right)$

$=\left(\frac{9}{2}, \frac{4}{2}\right)=\left(\frac{9}{2}, 2\right)$

$\therefore\left(\frac{3+a}{2}, \frac{1+b}{2}\right)=\left(\frac{9}{2}, 2\right)$

$\Rightarrow \frac{3+a}{2}=\frac{9}{2}$ and $\frac{1+b}{2}=2$

$\Rightarrow 3+a=9$ and $1+b=4$

$\Rightarrow a=6$ and $b=3$

Hence,  the values of a and b is 6 and 3, respectively.