1. Let the polynomials be

$f(x)=ax^2+bx+c \quad (a\neq 0), \qquad g(x)=px+q \quad (p\neq 0).$

We are given:

$f(g(x))=8x^2-2x$ and $g(f(x))=4x^2+6x+1.$

2. Use $f(g(x))$

Since $g(x)=px+q$,

$f(g(x))=a(px+q)^2+b(px+q)+c.$

Expanding,

$f(g(x))=ap^2x^2+(2apq+bp)x+(aq^2+bq+c).$

Compare with $8x^2-2x$:

$ap^2=8 \quad ...(1)$ $2apq+bp=-2 \quad ...(2)$ $aq^2+bq+c=0 \quad ...(3)$

3. Use $g(f(x))$

Since $g(y)=py+q$,

$g(f(x))=p(ax^2+bx+c)+q= pax^2+pbx+(pc+q).$

Compare with $4x^2+6x+1$:

$pa=4 \quad ...(4)$ $pb=6 \quad ...(5)$ $pc+q=1 \quad ...(6)$

4. Solve for $p,a,b$

From (1) and (4):

$ap^2=8, \qquad ap=4.$

Divide the first by the second:

$p=\frac{8}{4}=2.$

Then from $ap=4$,

$a=2.$

From (5):

$pb=6 \Rightarrow 2b=6 \Rightarrow b=3.$

5. Find $q$

Use (2):

$2apq+bp=-2.$

Substitute $a=2, p=2, b=3$:

$2\cdot 2\cdot 2\cdot q + 3\cdot 2 = -2$ $8q+6=-2$ $8q=-8$ $q=-1.$

So,

$g(x)=2x-1.$

6. Find $c$

From (6):

$pc+q=1.$

Substitute $p=2, q=-1$:

$2c-1=1$ $2c=2$ $c=1.$

Thus,

$f(x)=2x^2+3x+1.$

7. Compute $f(2)+g(2)$

$f(2)=2(2)^2+3(2)+1=8+6+1=15.$

$g(2)=2(2)-1=3.$

Therefore,

$f(2)+g(2)=15+3=18.$

8. Verification

Check quickly:

$f(g(x))=f(2x-1)=2(2x-1)^2+3(2x-1)+1$ $=2(4x^2-4x+1)+6x-3+1=8x^2-8x+2+6x-2=8x^2-2x.$

Correct.

$g(f(x))=2(2x^2+3x+1)-1=4x^2+6x+2-1=4x^2+6x+1.$

Also correct.

Hence the required value is indeed $18$.