Two sets A and B are as under: $$A = \{(a, b) \in R \times R : |a - 5| < 1$$ and $$|b - 5| < 1\}$$; $$B = \{(a, b) \in R \times R : 4(a - 6)^2 + 9(b - 5)^2 \leq 36\}$$. Then:

We have two subsets of the plane $$\mathbb R \times \mathbb R$$ defined as 

$$A=\{(a,b):|a-5|<1 \text{ and } |b-5|<1\}$$ and $$B=\{(a,b):4(a-6)^2+9(b-5)^2\le 36\}.$$

First we rewrite the description of each set in a form that is easier to interpret.

For set $$A$$, the inequalities $$|a-5|<1$$ and $$|b-5|<1$$ mean

$$4<a<6 \quad\text{and}\quad 4<b<6.$$

Thus $$A$$ is the open square whose centre is the point $$(5,5)$$ and whose side length is $$2$$.

For set $$B$$ we begin by recalling the standard form of an ellipse. The general formula

$$\frac{(x-h)^2}{p^2}+\frac{(y-k)^2}{q^2}\le 1$$

represents an ellipse centred at $$(h,k)$$ with semi-axes $$p$$ (along the $$x$$-direction) and $$q$$ (along the $$y$$-direction).

We rewrite the defining inequality of $$B$$ in this standard form. Starting from

$$4(a-6)^2+9(b-5)^2\le 36,$$

we divide every term by $$36$$:

$$\frac{4(a-6)^2}{36}+\frac{9(b-5)^2}{36}\le 1.$$

Simplifying the fractions, we obtain

$$\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}\le 1.$$

Hence $$B$$ is an ellipse with centre $$(6,5),$$ semi-major axis $$3$$ along the $$a$$-direction and semi-minor axis $$2$$ along the $$b$$-direction.

Now we test whether every point of $$A$$ also lies in $$B$$. So let us take an arbitrary point $$(a,b)\in A$$. By membership in $$A$$ we know

$$|a-5|<1 \; \Longrightarrow \; 4<a<6,$$

$$|b-5|<1 \; \Longrightarrow \; 4<b<6.$$

We next estimate the two squared distances that appear in the ellipse inequality.

Because $$4<a<6,$$ we have

$$|a-6|=6-a<6-4=2,$$

so

$$(a-6)^2<2^2=4.$$

Similarly, since $$4<b<6,$$ we get

$$|b-5|<1 \;\Longrightarrow\; (b-5)^2<1.$$

We substitute these upper bounds into the left-hand side of the ellipse condition:

$$\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}\;<\;\frac{4}{9}+\frac{1}{4}.$$

To compare this sum with $$1,$$ we find a common denominator $$36$$:

$$\frac{4}{9}+\frac{1}{4}=\frac{16}{36}+\frac{9}{36}=\frac{25}{36}.$$

Because $$\frac{25}{36}<1,$$ we have

$$\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}<1.$$

This strict inequality certainly implies the weak inequality

$$\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}\le 1,$$

which is exactly the condition for $$(a,b)$$ to belong to $$B$$. Therefore every point of $$A$$ lies inside the ellipse $$B$$; symbolically, $$A\subset B.$$

It remains to see that the reverse inclusion does not hold. Consider, for instance, the centre of the ellipse, the point $$P=(6,5).$$ Substituting into the ellipse equation gives

$$4(6-6)^2+9(5-5)^2=0\le 36,$$

so $$P\in B.$$ However, for this point we have $$|a-5|=|6-5|=1,$$ which is not strictly less than $$1,$$ hence $$P\notin A.$$ Thus $$B\not\subset A.$$

Combining these results we conclude

$$A\subset B \quad\text{but}\quad B\not\subset A.$$

Hence, the correct answer is Option C.