Solution to Part (a). One can establish this in several ways. For instance, using the convexity of $t \to t^2$ we have
$$\left(\frac{3+2(x+y+z)}{3}\right)^2 = \left(\frac{(1+x+y)+(1+y+z)+(1+z+x)}{3}\right)^2 \le \frac{(1+x+y)^2+(1+y+z)^2+(1+z+x)^2}{3},$$
with equality iff
$$1+x+y=1+y+z=1+z+x,$$
i.e., $x = y = z$. Hence
$$\frac{(3+2(x+y+z))^2}{3} \le (1+x+y)^2 + (1+y+z)^2 + (1+z+x)^2,$$
with equality iff $x = y = z$.
This also follows from an application of the Cauchy-Schwarz inequality
to the expression
$$3+2(x+y+z) = (1+x+y) + (1+y+z) + (1+z+x).$$
To continue, by the AM-GM inequality,
$$3+2(x+y+z) \ge 3+6\sqrt{xyz} \ge 9,$$
with equality iff $x = y = z = 1$. Hence
$$27 = \frac{81}{3} \le \frac{(3+2(x+y+z))^2}{3} \le (1+x+y)^2 + (1+y+z)^2 + (1+z+x)^2,$$
with equality iff $x = y = z$.
Alternatively, applying the AM-GM inequality termwise we see that
$$\begin{aligned} (1+x+y)^2 + (1+y+z)^2 + (1+z+x)^2 &\ge (3\sqrt[3]{1xy})^2 + (3\sqrt[3]{1yz})^2 + (3\sqrt[3]{1zx})^2 \\ &= 9(\sqrt[3]{x^2y^2} + \sqrt[3]{y^2z^2} + \sqrt[3]{z^2x^2}) \\ &\ge 27\sqrt[3]{x^4y^4z^4} \\ &\ge 27, \end{aligned}$$
with equality throughout iff $x = y = z = 1$.

Solution to Part (b). Now consider the inequality on the right-hand side:
$$(1+x+y)^2 + (1+y+z)^2 + (1+z+x)^2 \le 3(x+y+z)^2.$$
Expanding and simplifying this leads to the following equivalent statement that
$$3+2(x^2+y^2+z^2)+2(xy+yz+zx)+4(x+y+z) \le 3(x^2+y^2+z^2)+6(xy+yz+zx),$$
i.e.,
$$3+4(x+y+z) \le x^2+y^2+z^2+4(xy+yz+zx) = (x+y+z)^2+2(xy+yz+zx).$$
Equivalently,
$$7 \le (u-2)^2 + 2v,$$
where $u = x + y + z$, $v = xy + yz + zx$. But $u \ge 3$ and $v \ge 3$, with equality in both of these inequalities iff $x = y = z = 1$. Hence
$$(u-2)^2 + 2v \ge 1+6=7,$$
with equality iff $x = y = z = 1$. This establishes the desired result.