For the first question:

(a) On $\Bbb C^3$ choose the basis $e_1, e_2, e_3$ and then let $\sigma\in S^3$ act on an element $v=\sum_i v_i e_i$ by $\sigma\cdot v = \sum_i v_i e_{\sigma(i)}$. This is a linear representation of $S^3$ on $\Bbb C^3$ and it admits a two dimensional subrepresentation given by vectors of the form $\lambda_1 \cdot (e_1-e_2)+\lambda_2\cdot (e_2-e_3)$.

This subrepresentation is called the standard representation. For the character one calculates

$$Tr(1) = 2$$

since the representation is two-dimensional. One checks $(12)\cdot (e_1-e_2)=-(e_1-e_2) $, and $(12)\cdot (e_2-e_3)=(e_1-e_2)+(e_2-e_3)$ whence

$$Tr((12))=-1+1=0$$

Further: $(123)\cdot (e_1-e_2) = (e_2-e_3)$ and $(123)\cdot (e_2-e_3) = -(e_1-e_2)-(e_2-e_3)$

Whence

$$Tr((123))=-1$$

This computes the character since the conjugacy clases of $S_3$ generated by the elements $1, (12), (123)$.

(b) For two representations $V,W$ we have $Hom(V,W) \cong V^*\otimes W$ so that $Hom(V_{std}, Hom(V_{std},V_{std})) \cong V_{std}^*\otimes V_{std}^*\otimes V_{std}$.

Further one has the rules $\chi_{V\otimes W}(\sigma)= \chi_{V}(\sigma)\cdot \chi_{W}(\sigma)$ and $\chi_{V^*}(\sigma)=\chi_{V}(\sigma^{-1})$.

Letting $\tau$ denote the character of the representation $Hom(V_{std}, Hom(V_{std},V_{std}))$ this gives

$$\tau(1) = 2^3=8,\qquad \tau((12))=0^3=0,\qquad \tau((123)) = (-1)^3=(-1)$$

(recall that the inverse of $(12)$ is $(12)$ and that the inverse of $(123)$ is $(123)^2=(312)$ which lies in the same conjugacy class).

If we let $\chi_{std}, \chi_{sign}, \chi_{triv}$ denote the characters of the standard, sign and trivial representations we see:

$$\tau=3\cdot \chi_{std}+\chi_{sign}+\chi_{triv}$$

(since $\chi_{sign}$ applied to $1, (12), (123)$ gives $1,-1,1$ and $\chi_{triv}$ is constant $1$).

So $Hom(V_{std}, Hom(V_{std},V_{std}))$ has the same character as $V_{std}\oplus V_{std}\oplus V_{std}\oplus V_{triv}\oplus V_{sign}$.

By the characterisation of finite dimensional representations of finite groups via their characters (I don't know the name of the theorem) one finds that $Hom(V_{std}, Hom(V_{std},V_{std}))$ is then isomorphic to the above representation.

(c) Because of the remark about $(12)^{-1}=(12)$ and $(123)^{-1}$ having the same conjugacy class as $(123)$ one sees that for any finie dim. representation $V$ of $S^3$ the character of $V$ and $V^*$ agree. Hence $V^*$ and $V$ are isomorphic.

Then

$$Hom(V_{std}, Hom(V_{std},V_{std}))\cong V_{std}^*\otimes V_{std}^*\otimes V_{std} \cong V_{std}\otimes V_{std} \otimes V_{std}$$

and we have already found a decomposition of the left-hand side in the previous exercise.

(d) If a representation has dimension $5$ then we have $\chi(1)=5$. By the fundamental theorem of fintie dimensional representations of finite groups we must be able to write:

$$\chi = n_1 \chi_{triv}+n_2\chi_{sign}+n_3\chi_{std}$$

where $n_1,n_2,n_3\ge0$ are integers. The condition $\chi(1)=5$ becomes $n_1+n_2+2n_3=5$. The condition $\chi((13))=1$ becomes $n_1-n_2=1$. The condition $\chi(123)=0$ becomes $n_1+n_2-n_3=0$.

The last equation allows us to read off $n_3=n_1+n_2$, putting it into the first equation gives $3(n_1+n_2)=5$, which is impossible since $5$ is not divisble by $3$. So no such representation may exist.

For the second question:

(a) The definition of the induced representation is

$$Ind^G_K(W) = W\otimes_{\Bbb C[K]} \Bbb C[G]$$

where $\Bbb C[K]$ is the group-ring of $K$. Using that $\Bbb C[H]\otimes_{\Bbb C[H]} \Bbb C[G] = \Bbb C[G]$ if $H$ is a sub-group of $G$ one gets:

$$Ind_H^G(Ind_K^H(W)) = (W\otimes_{\Bbb C[K]} \Bbb C[H])\otimes_{\Bbb C[H]} \Bbb C[G] \cong W \otimes_{\Bbb C[K]}(\Bbb C[H]\otimes_{\Bbb C[H]}\Bbb C[G])\\ \cong W\otimes_{\Bbb C[K]}\Bbb C[G] = Ind_K^G(W)$$

(b) The formula for the character of an induced representation is as follows:

$$\chi_{Ind_H^G(V)}(g) = \frac1{|H|}\sum_{x\in G: x^{-1}gx\in H}\chi_{V}(x^{-1}gx)$$

Now $H$ is the subgroup generated by $y$. Since $y^4=1$ one has $|H|=4$.

Because of $yx = (xx)yx(yy^3)=x(xy)(yx)y^3)=xy^3$ you can always bring any element of $D$ into the form $x^ky^l$ where $k\in\{0,1\}, l\in\{0,1,2,3\}$. The group $D_8$ is then given by the elements $1, x, y, y^2,y^3, xy, xy^2,xy^3$, which are all distinct.

These elements further fall into $5$ conjugacy classes:

$1$ - which only has the identity

$\{x, xy^2\}$ - (you have $yxy^3 = xy^6 = xy^2$)

$\{y, y^3, \}$ - (you have $xyx=xxy^3=y^3$)

$\{y^2\}$

$\{xy, xy^3\}$ - (you have $xxyx=xy^3$)

This means that only the conjugacy classes of $1$, $y$, $y^2$ in $G$ have elements in $H$ - for all other conjugacy classes the character will then evaluate to $0$. Infact one has that every element of these conjugacy classes is in $H$! Using that $\chi_V(g)=\dim(V)$ for all elements $g$ of $H$ allows us to now directly apply the character formula to get:

$\chi_{Ind_H^G(V)}(1) = \chi_{Ind_H^G(V)}(y) = \chi_{Ind_H^G(V)}(y^3) = \frac14 \cdot 8\cdot \dim(V)=2\dim(V)$

$\chi_{Ind_H^G(V)}(x) = \chi_{Ind_H^G(V)} (xy)= 0 $

I now assume that $\dim(V)=1$ since the exercise mentions "the" trivial representation rather than just "a" trivial representation. It is easy to check that the map $G\to \Bbb C^*$ that sends $x^ky^l$ to $(-1)^k$ is a group homomorphism and hence a $1d$ representation of $G$. Call this representation $V_{x}$.

Then $\chi_{V_x}(1)=\chi_{V_x}(y)=\chi_{V_x}(y^2)=1$ and $\chi_{V_x}(x)=\chi_{V_x}(xy)=-1$ as is easy to check.

Denoting with $V_{trvG}$ the trivial representation of $G$ on $\Bbb C$ one then gets:

$$\chi_{Ind_H^G(V)} = \chi_{V_x}+\chi_{V_{trvG}}$$

So $Ind_H^G(V) \cong V_x\oplus V_{trvG}$.