If $$I_1 = \int_0^1 e^{-x} \cos^2 x \, dx$$; $$I_2 = \int_0^1 e^{-x^2} \cos^2 x \, dx$$ and $$I_3 = \int_0^1 e^{-x^2} dx$$; then:

We have three definite integrals defined on the common interval $$[0,1]$$:

$$I_1=\displaystyle\int_0^1 e^{-x}\,\cos^2x\,dx,$$

$$I_2=\displaystyle\int_0^1 e^{-x^2}\,\cos^2x\,dx,$$

$$I_3=\displaystyle\int_0^1 e^{-x^2}\,dx.$$

To decide their relative magnitudes, we shall compare their integrands point-wise for every $$x$$ in $$[0,1]$$ and then use the fact that if for every $$x$$ in the interval $$f(x)<g(x)$$, then the definite integrals satisfy $$\int f<\int g$$ over that interval.

Step 1: Comparing the exponential parts.

For any $$x$$ with $$0\le x\le1$$ we observe that $$x^2\le x$$. Multiplying by the negative sign reverses the inequality:

$$-x^2\ge -x.$$

Exponentiating by the natural base $$e$$ (and recalling that the exponential function is strictly increasing) preserves the inequality direction, giving

$$e^{-x^2}\ge e^{-x}.$$ Since $$x^2<x$$ for every $$x\in(0,1)$$, the inequality is actually strict there:

$$e^{-x^2}>e^{-x}\qquad\text{for }0<x\le1.$$ Thus, for every $$x\in[0,1]$$, we have

$$e^{-x^2}\ge e^{-x}\quad\text{and}\quad e^{-x^2}>e^{-x}\text{ whenever }x\in(0,1].$$

Step 2: Using the common factor $$\cos^2x$$.

The factor $$\cos^2x$$ satisfies $$0\le\cos^2x\le1$$ for all real $$x$$, so in particular for all $$x\in[0,1]$$ we have

$$0\le\cos^2x\le1.$$ Multiplying the inequality in Step 1 by this non-negative factor preserves the inequality:

$$e^{-x^2}\cos^2x\;\;\ge\;\;e^{-x}\cos^2x,$$ and the inequality is again strict for every $$x\in(0,1]$$ because both factors are positive there. Therefore, for the integrands we can write

$$\boxed{e^{-x^2}\cos^2x\;>\;e^{-x}\cos^2x\quad\text{for }0<x\le1.}$$

Step 3: Comparing $$I_2$$ with $$I_3$$.

Notice that

$$e^{-x^2}\cos^2x = e^{-x^2}\times\cos^2x,$$ and since $$\cos^2x\le1,$$ we obviously have

$$\boxed{e^{-x^2}\cos^2x\;\le\;e^{-x^2}\quad\text{for all }x\in[0,1],}$$ with strict inequality except at those points where $$\cos^2x=1$$ (that occurs only at $$x=0$$ within $$[0,1]$$).

Step 4: Translating point-wise inequalities into integral inequalities.

Because the definite integral over a fixed interval preserves inequality when the integrand inequality holds everywhere on that interval, we integrate the boxed inequalities obtained above.

First, integrating $$e^{-x^2}\cos^2x \le e^{-x^2}$$ from $$0$$ to $$1$$ gives

$$\displaystyle\int_0^1 e^{-x^2}\cos^2x\,dx\;\le\;\int_0^1 e^{-x^2}\,dx,$$

that is,

$$I_2\;\le\;I_3.$$

The inequality is strict because the integrands differ on a set of positive measure (indeed everywhere except possibly at $$x=0$$), so we actually have

$$\boxed{I_2\;<\;I_3.}$$

Next, integrating $$e^{-x^2}\cos^2x > e^{-x}\cos^2x$$ from $$0$$ to $$1$$ yields

$$\displaystyle\int_0^1 e^{-x^2}\cos^2x\,dx\;>\;\int_0^1 e^{-x}\cos^2x\,dx,$$

that is,

$$\boxed{I_2\;>\;I_1.}$$

Step 5: Assembling the chain of inequalities.

Combining the two boxed results, we obtain the single strict chain

$$I_3\;>\;I_2\;>\;I_1.$$

This inequality exactly matches Option 4 in the given list.

Hence, the correct answer is Option 4.