解: (1)当\( a = -1 \)时,\( f(x) = x^2 + 2\ln x + 1 \),

则\( f'(x) = 2x + \displaystyle \frac{2}{x} \),所以\( f'(1) = 4 \),

又\( f(1) = 2 \),所以曲线\( y = f(x) \)在\( x = 1 \)处的切线方程为\( y = 4(x – 1) + 2 \),即\( y = 4x – 2 \),

联立方程\( \begin{cases} y = 4x – 2 \\ y = x^2 + 2\ln x + 1 \end{cases} \)，可得\( x^2 + 2\ln x – 4x + 3 = 0 \),

令\( g(x) = x^2 + 2\ln x – 4x + 3, x > 0 \),

则\( g'(x) = 2x + \displaystyle \frac{2}{x} – 4 \geq 0 \),

所以\( g(x) \)在\( (0, +\infty) \)上单调递增.

又因为\( g(1) = 0 \),所以\( g(x) = 0 \)有且仅有1解,

所以直线\( l \)与曲线\( y = f(x) \)公共点个数为1；

(2)\( \forall x_1, x_2 \in [1, e], |f(x_1) – f(x_2)| < e^2 + 1 \)恒成立,

等价于\( f(x)_{\max} – f(x)_{\min} < e^2 + 1 \),

由\( f(x) = x^2 – 2a\ln x + 1 \),得\( f'(x) = 2x – \displaystyle \frac{2a}{x} = \displaystyle \frac{2(x^2 – a)}{x}, a > 0, x > 0 \),

当\( x \in (0, \sqrt{a}) \)时,\( f'(x) < 0 \),所以\( f(x) \)在\( (0, \sqrt{a}) \)上单调递减；

当\( x \in (\sqrt{a}, +\infty) \)时,\( f'(x) > 0 \),所以\( f(x) \)在\( (\sqrt{a}, +\infty) \)上单调递增,

①当\( \sqrt{a} \leq 1 \),即\( 0 < a \leq 1 \)时,\( f(x) \)在\( [1, e] \)上单调递增.

所以\( f(x)_{\min} = f(1) = 2, f(x)_{\max} = f(e) = e^2 – 2a + 1 \),此时\( e^2 – 2a + 1 – 2 = e^2 – 2a – 1 < e^2 + 1 \),所以\( 0 < a \leq 1 \)满足条件；

②当\( \sqrt{a} \geq e \),即\( a \geq e^2 \)时,\( f(x) \)在\( [1, e] \)上单调递减, 所以\( f(x)_{\max} = f(1) = 2, f(x)_{\min} = f(e) = e^2 – 2a + 1 \), 此时\( 2 – e^2 + 2a – 1 = 2a – e^2 + 1 \geq e^2 + 1 \),

不合题意,

③当\( 1 < \sqrt{a} < e \),即\( 1 < a < e^2 \)时,\( f(x) \)在\( (1, \sqrt{a}) \)上单调递减,在\( (\sqrt{a}, e) \)上单调递增,

所以\( f(x)_{\min} = f(\sqrt{a}) = a – a\ln a + 1, f(x)_{\max} = \max\{f(1), f(e)\} \),

所以\( 2 – (a – a\ln a + 1) < e^2 + 1 \),且\( (e^2 - 2a + 1) - (a - a\ln a + 1) < e^2 + 1 \),即\( a\ln a - a - e^2 < 0 \),且\( a\ln a - 3a - 1 < 0 \),

令\( m(a) = a\ln a – a – e^2 \),当\( 1 < a < e^2 \)时,\( m'(a) = \ln a > 0 \),所以\( m(a) \)在\( (1, e^2) \)上单调递增,所以\( m(a) < m(e^2) = 0 \),

令\( t(a) = a\ln a – 3a – 1 \),当\( 1 < a < e^2 \)时,\( t'(a) = \ln a - 2 < 0 \), 所以\( t(a) \)在\( (1, e^2) \)上单调递减,所以\( t(a) < t(1) = -4 < 0 \),

所以当\( 1 < a < e^2 \)时,满足题意.

综上, \( a \)的取值范围为\( a \in (0, e^2) \)。

（1）\( g(x) = f'(x) = (1 – a)\sin x + x\cos x \)

\( g'(x) = (2 – a)\cos x – x\sin x \)

因为\( x = \pi \)是\( g(x) \)的极小值点，

所以\( g'(\pi) = 0 \)，所以\( a = 2 \)

此时，\( g'(x) = -x\sin x \)

经验证，\( x = \pi \)是\( g(x) \)的极小值点。

所以\( a = 2 \)

（2）\( f(x) = x\sin x + 2\cos x \)，\( x \in (-2\pi, 2\pi) \)

因为\( f(-x) = f(x) \)，所以\( f(x) \)为偶函数

先考虑\( x \in [0, 2\pi) \)

当\( x = \displaystyle\frac{\pi}{2} \)或\( x = \displaystyle\frac{3\pi}{2} \)时，\( f(x) \neq 0 \)。

当\( x \neq \displaystyle\frac{\pi}{2} \)且\( x \neq \displaystyle\frac{3\pi}{2} \)时，

令\( f(x) = x\sin x + 2\cos x = 0 \)

所以\( \tan x = -\displaystyle\frac{2}{x} \)，由图象可知有2个交点，

所以\( x \in (0, 2\pi) \)时\( f(x) \)有2个零点，

因为\( f(x) \)为偶函数，所以\( x \in (-2\pi, 0) \)时，\( f(x) \)有2个零点

所以\( f(x) \)共有4个零点。