A circuit contains three capacitors in series: 3μF, 6μF, and 2μF. Another branch has two capacitors in series: 4μF and 4μF. These two branches are in parallel. What is the total capacitance?

The capacitors are $3\mu\text{F}$, $6\mu\text{F}$, and $2\mu\text{F}$ in series.
For capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances.

$$\frac{1}{C_{\text{top}}} = \frac{1}{3\mu\text{F}} + \frac{1}{6\mu\text{F}} + \frac{1}{2\mu\text{F}}$$ $$\frac{1}{C_{\text{top}}} = \frac{2}{6\mu\text{F}} + \frac{1}{6\mu\text{F}} + \frac{3}{6\mu\text{F}}$$ $$\frac{1}{C_{\text{top}}} = \frac{2+1+3}{6\mu\text{F}} = \frac{6}{6\mu\text{F}} = \frac{1}{1\mu\text{F}}$$ $$C_{\text{top}} = 1\mu\text{F}$$

The capacitors are $4\mu\text{F}$ and $4\mu\text{F}$ in series.

$$\frac{1}{C_{\text{bottom}}} = \frac{1}{4\mu\text{F}} + \frac{1}{4\mu\text{F}}$$ $$\frac{1}{C_{\text{bottom}}} = \frac{2}{4\mu\text{F}} = \frac{1}{2\mu\text{F}}$$ $$C_{\text{bottom}} = 2\mu\text{F}$$

For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances.

$$C_{\text{parallel}} = C_{\text{top}} + C_{\text{bottom}}$$ $$C_{\text{parallel}} = 1\mu\text{F} + 2\mu\text{F}$$ $$C_{\text{parallel}} = 3\mu\text{F}$$

The $2\mu\text{F}$ capacitor on the left, $C_{\text{parallel}}$, and the $2\mu\text{F}$ capacitor on the right are all in series.

$$\frac{1}{C_{\text{effective}}} = \frac{1}{2\mu\text{F}} + \frac{1}{C_{\text{parallel}}} + \frac{1}{2\mu\text{F}}$$ $$\frac{1}{C_{\text{effective}}} = \frac{1}{2\mu\text{F}} + \frac{1}{3\mu\text{F}} + \frac{1}{2\mu\text{F}}$$

Find a common denominator, which is $6\mu\text{F}$.

$$\frac{1}{C_{\text{effective}}} = \frac{3}{6\mu\text{F}} + \frac{2}{6\mu\text{F}} + \frac{3}{6\mu\text{F}}$$ $$\frac{1}{C_{\text{effective}}} = \frac{3+2+3}{6\mu\text{F}} = \frac{8}{6\mu\text{F}} = \frac{4}{3\mu\text{F}}$$ $$C_{\text{effective}} = \frac{3}{4}\mu\text{F}$$ $$C_{\text{effective}} = 0.75\mu\text{F}$$

The effective capacitance is $0.75\mu\text{F}$. This matches option B.

The final answer is $\boxed{\text{0.75\mu F}}$.