Let $$f$$ be a twice differentiable function on $$\mathbb{R}$$. If $$f'(0) = 4$$ and $$f(x) + \displaystyle\int_0^x (x-t)f'(t) dt = (e^{2x} + e^{-2x})\cos 2x + \dfrac{2}{a}x$$, then $$(2a+1)^5a^2$$ is equal to ______.

Given that $$f$$ is twice differentiable on $$\mathbb{R}$$, $$f'(0) = 4$$, and:

$$f(x) + \int_0^x (x-t)f'(t)\,dt = (e^{2x} + e^{-2x})\cos 2x + \frac{2}{a}x \quad \cdots (*)$$

Simplify the integral using integration by parts on $$\int_0^x t f'(t)\,dt$$:

$$\int_0^x (x-t)f'(t)\,dt = x\int_0^x f'(t)\,dt - \int_0^x t f'(t)\,dt$$

$$= x[f(x) - f(0)] - \left[xf(x) - \int_0^x f(t)\,dt\right] = \int_0^x f(t)\,dt - xf(0)$$

So equation (*) becomes:

$$f(x) + \int_0^x f(t)\,dt - xf(0) = (e^{2x} + e^{-2x})\cos 2x + \frac{2x}{a} \quad \cdots (1)$$

Setting $$x = 0$$ in equation (1) gives:

$$f(0) + 0 - 0 = (1 + 1)(1) + 0 = 2$$

$$\therefore f(0) = 2$$

Differentiate equation (1) with respect to $$x$$:

$$f'(x) + f(x) - f(0) = \frac{d}{dx}\left[(e^{2x} + e^{-2x})\cos 2x\right] + \frac{2}{a}$$

Let $$g(x) = (e^{2x} + e^{-2x})\cos 2x$$. Then:

$$g'(x) = (2e^{2x} - 2e^{-2x})\cos 2x - 2(e^{2x} + e^{-2x})\sin 2x$$

At $$x = 0$$ we have:

$$g'(0) = (2 - 2)(1) - 2(1 + 1)(0) = 0$$

Substituting $$x = 0$$ into the differentiated equation yields:

$$f'(0) + f(0) - f(0) = g'(0) + \frac{2}{a}$$

$$4 = 0 + \frac{2}{a}$$

$$a = \frac{1}{2}$$

Compute the final expression $$(2a + 1)^5 \cdot a^2$$:

$$(2a + 1)^5 \cdot a^2 = \left(2 \cdot \frac{1}{2} + 1\right)^5 \cdot \left(\frac{1}{2}\right)^2 = 2^5 \cdot \frac{1}{4} = 32 \times \frac{1}{4} = 8$$

The answer is $$\boxed{8}$$.