a. In order for
to be a proper density function, its integral over its domain must evaluate to 1:

also must be non-negative over its support, which is the case here.
I assume the instruction regarding the distribution function doesn't apply to parts (b) and (c).
(b) Integrate the density over the interval [0.1, 0.2]:

(c) Integrate the density over the interval [0.5, 1]:

The distribution function is obtained by integrating the density:

(d) Using the distribution function, we have

(e) Using
again, we get

(f) The mean is
![E[X]=\displaystyle\int_(-\infty)^\infty x\,f(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13](https://img.qammunity.org/2021/formulas/mathematics/college/s5d5jzcpbodbv77zeaoe1sq2kakk1hcxjt.png)
The variance is
![V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2](https://img.qammunity.org/2021/formulas/mathematics/college/e2e5hg9sy4v38tj5q4nl365utlyba8oawi.png)
where
![E[X^2]=\displaystyle\int_(-\infty)^\infty x^2\,f(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16](https://img.qammunity.org/2021/formulas/mathematics/college/ylcy9cpbt9li0vri82hql39m3h3czxi3tj.png)
so that the variance is
![V[X]=\frac16-\left(\frac13\right)^2=\frac1{18}](https://img.qammunity.org/2021/formulas/mathematics/college/irqsm9lbg5ywec953bp5ktwcyaylf5xd1z.png)