Ans: (a) $-4 \le x < 7$ (b) $6$
- Consider the first inequality, we have
$\begin{array}{rcl}
\dfrac{7(x-2)}{5}+11 & > & 3(x-1) \\
7x-14 + 55 & > & 15x-15 \\
-8x & > & -56 \\
x & < & 7 \end{array}$Consider the second inequality, we have
$\begin{array}{rcl}
x + 4 & \ge & 0 \\
x & \ge & -4
\end{array}$
Therefore, the overall solutions are $-4 \le x <7$.
- There are $6$ positive integers satisfying the both inequalities.
