For part B, you are correct as far as you've gone. There is no algebraic solution to your equation. A graphing calculator or Newton's Method iteration can get you to a solution fairly quickly.
a ≈ 1.114157141
For part C, you need to consider your answer. For a=0, the equation is that of a straight line, so there is no inflection point at x=1. For cos(a)=0, there are an infinite number of possible values of
![a](https://img.qammunity.org/2019/formulas/mathematics/middle-school/29wl7w45fjbhkymu53hsles9qvadja8y1s.png)
that will put a point of inflection at x=1. As you have noted, a=π/2 is only one of them in the range 0 < a < 4.
For part D, again you have stopped part way to the answer. Consider what values of
![a](https://img.qammunity.org/2019/formulas/mathematics/middle-school/29wl7w45fjbhkymu53hsles9qvadja8y1s.png)
will make
![a \sin(ax)](https://img.qammunity.org/2019/formulas/mathematics/high-school/zhl1gh2ssa0pedmufipaxna7chuqjpe5yg.png)
strictly greater than 1. There aren't any. The sine function always crosses zero. This part of the question has no solution.