We can use the Bernoulli equation and continuity equation to solve this problem. The Bernoulli equation is given by:

$$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$$

The continuity equation is given by:

$$A_1 v_1 = A_2 v_2$$

From the given data, we have:

$$P_1 - P_2 = 3 \mathrm{~Nm}^{-2}$$ $$A_1 = 10 \mathrm{~cm}^2 = 10 \times 10^{-4} \mathrm{~m}^2$$ $$A_2 = 5 \mathrm{~cm}^2 = 5 \times 10^{-4} \mathrm{~m}^2$$ $$\rho = 1.25 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$$

Rearrange the continuity equation to solve for $$v_2$$:

$$v_2 = \frac{A_1}{A_2} v_1 = 2v_1$$

Substitute $$v_2$$ and rearrange the Bernoulli equation:

$$P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$$

$$3 = \frac{1}{2} \times 1.25 \times 10^3 (4v_1^2 - v_1^2)$$

Now, solve for $$v_1$$:

$$3 = \frac{1}{2} \times 1.25 \times 10^3 \times 3v_1^2$$ $$v_1^2 = \frac{3}{1.875 \times 10^3}$$

$$v_1 = \sqrt{\frac{3}{1.875 \times 10^3}}$$

$$v_1 \approx 0.0400 \mathrm{~m} \mathrm{~s}^{-1}$$

Now, calculate the rate of flow of glycerin through the pipe (volume flow rate) using $$v_1$$ and $$A_1$$:

$$Q = A_1 v_1$$

$$Q = 10 \times 10^{-4} \times 0.0400$$

$$Q = 4 \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$$

So, the rate of flow of glycerin through the pipe is $$4 \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$$, and the value of $$x$$ is 4.