Answer
a. $b=-m\pi$ or $m=-\frac{b}{\pi}$ .
b. $m=-1$. $b=\pi$
Work Step by Step
a. For the function $f(x)=\begin{cases} sin( x),\ x\lt\pi \\ mx+b,\ x\geq\pi \end{cases}$ to be continuous at $x=\pi$, let $\lim_{x\to\pi^-}f(x)=\lim_{x\to\pi^+}f(x)$ or $\lim_{x\to\pi^-}sin(x)=m\pi +b$
We have $b=-m\pi$ or $m=-\frac{b}{\pi}$ (there are many sets of $m, b$ values in this case).
b. For the function to be differentiable at $x=\pi$, we need $\lim_{x\to\pi^-}f'(x)=\lim_{x\to\pi^+}f'(x)$ or $\lim_{x\to\pi^-}cos(x)=m$ which gives $m=-1$. For the function to satisfy both conditions, we have $b=\pi$.
