I will just give the solution for 2. (a) since the remaining case is very similar.
Clearly, if $\displaystyle f$ is increasing then we know that $\displaystyle f(x)\geq f(a)$ for all $\displaystyle x\in[a,b]$.
Hence, we have
$\displaystyle \int\limits_{a}^{b}f(x)dx\geq\int\limits_{a}^{b}f( a)dx=f(a)(b-a),$
which completes the first part of the inequality.
For the remaining part, we consider that $\displaystyle f$ is concave down (convex).
This indicates that $\displaystyle f(x)\leq\frac{f(b)-f(a)}{b-a}(x-a)+f(a)$ holds for all $\displaystyle x\in[a,b]$ (draw a graph for the increasing convex function $\displaystyle f$ and the line passing at the points $\displaystyle (a,f(a))$ and $\displaystyle (b,f(b))$).
Integrating both sides of this inequality, we get
$\displaystyle \int\limits_{a}^{b}f(x)dx\leq\int\limits_{a}^{b}\B igg(\frac{f(b)-f(a)}{b-a}(x-a)+f(a)\Bigg)dx=(b-a)\frac{f(b)+f(a)}{2},$
which completes the proof.