1, not clear.

(4)x→0 The original formula =-2/(-cosx) = 2.

3,f(x)=(x-3)(x-8)^(2/3)，

f′(x)=(x-8)^(2/3)+(2/3)(x-3)(x-8)^(- 1/3)。

4. The original formula = 2 [(2e 2x) sin (2x)+2 * 2 * 3 (e 2x) sin? (2x)cos(2x)]|0→π

=0。