Solution:
$x(t)$ is periodic with period $=3$, so $\omega_0=\frac{2\pi}{3}$.
$x(t)$ is even, so its Fourier series expansion has cosines only.

Brick-wall lowpass filter with cutoff frequency 1 radians/s

Solution:
Both $\pi/3$ and $2\pi/3>1$. But there is clearly a nonzero average value $a_0$.

$\displaystyle h(t)=\frac{\sin(0.4t)}{\pi t}\cos(0.5t)$

Solution:
$h(t)$ is a bandpass filter that passes only
$\displaystyle -0.4+0.5=0.1<|\omega|<0.4+0.5=0.9<1. $
This excludes $\omega=0$ and both $\pi/3$ and $2\pi/3>1$.

${\bf H} (\omega)=1-e^{-j3\omega}$

Solution:
All you need to know is that the period of $x(t)$ is 3. Then
$\displaystyle {\bf Y} (\omega)={\bf H} (\omega)\;{\bf X} (\omega)={\bf X} (\omega)-e^{-j3\omega}\;{\bf X} (\omega). $
Then ${\cal F}^{-1}\to y(t)=x(t)-x(t-3)=0$!

$\displaystyle h(t)=\frac{\sin(12t)}{\pi t}$

Solution:
Period $=0.5$ s$\to$ harmonics $2,4,6,\dots{\rm Hz}=4\pi,8\pi,12\pi,\dots$ radians/s.
${\bf H} (\omega)=1$ for $|\omega|<12<4\pi$; 0 otherwise. Only constant ($\omega=0$) term $c_0$ passes.

${\bf H} (\omega)=\cases{j\omega &for $2<|\omega|<4$\cr 0 &otherwise}$

Solution:
${\bf H} (\omega)$ eliminates all but the second term, and differentiates it. Output: $\cos(t)$.

Solution:
The modulated signal has components at $\pm 9\pm 0,2,4=\pm\{5,7,9,11,13\}$.
Only the components at $\pm 5$ get through the filter.

$h(t)=1,$ -2<t<2 and 0 otherwise

Solution:
$\displaystyle {\bf H} (\omega) = {\cal F}\{h(t)\}=4\frac{\sin(2\omega)}{2\omega}, $
$\displaystyle {\bf H}(\pi/4) = 4\frac{\sin(\pi/2)}{\pi/2}=\frac{8}{\pi}\;. $
Output is:
$\displaystyle\frac{8}{\pi}\cos(\frac{\pi}{4}t)$.

$\displaystyle y(t)=\int_{t-6}^tx(\tau)\;d\tau$

Solution:
The system has $h(t)=1$ for 0<t<6 and 0 otherwise. To find ${\bf H} (\omega)$:
Let $\tilde h(t)=1$ with $|t|<3$ or 0 otherwise. Then from Eq. (5.88), ${\bf \tilde H}(\omega)=6{{\sin(3\omega)}\over{3\omega}}$
Then $h(t)=\tilde h(t-3)$ and ${\bf H} (\omega)=6{{\sin(3\omega)}\over{3\omega}}e^{-j3\omega}$ and ${\bf H} (\pi/3)=0$!
The reason: the system integrates the cosine over a period, and this integral is zero.

Solution:
Let $x(t)=\frac{\sin(\pi t)}{\pi t}$.
Then ${\bf X} (\omega)=1$ for $|\omega|$<$\pi$ and 0 otherwise.
By Parseval's theorem,
$\displaystyle \int_{-\infty}^{\infty}|x(t)|^2\;dt =\frac1{2\pi}\;\int_{-\infty}^{\infty}|{\bf X} (\omega)|^2\;d\omega= \frac1{2\pi}\;\int_{-\pi}^{\pi}|1|^2\;d\omega=1! $

Solution:
$\displaystyle {\cal F}\{x(t)\} = 1{\rm for }|\omega|<2, $
$\displaystyle {\cal F}\left\{\frac{d^2x}{dt^2}\right\} = -\omega^2{\rm for }|\omega|<2. $
By Parseval's theorem,
$\displaystyle {\rm energy}={1\over{2\pi}} \;\int_{-2}^2\left|-\omega^2\right|^2\;d\omega =\frac{32}{5\pi}\;. $

Solution:
$\displaystyle 8j\omega\frac{\sin(4\omega)}{4\omega} = 2j\sin(4\omega)=e^{j4\omega}-e^{-j4\omega}, $
$\displaystyle {\cal F}^{-1}\left\{e^{j4\omega}-e^{-j4\omega}\right\} = \delta(t+4)-\delta(t-4). $

Method #2:
$\displaystyle {\cal F}^{-1}\left\{8\frac{\sin(4\omega)}{4\omega}\right\} = {\rm rect} \left(\frac{t}{8}\right)\;, $
$\displaystyle {\cal F}^{-1}\left\{8j\omega\frac{\sin(4\omega)}{4\omega}\right\} = \frac{d}{dt}\;{\rm rect}\left(\frac{t}{8}\right)=\delta(t+4)-\delta(t-4). $

Solution:
$e^{-a|t|}=e^{-at}\;u(t)+e^{at}u(-t)$.
Entries #7a and #7b in Table 5-6 are:
$\displaystyle {\cal F}\{e^{-at}\;u(t)\}=\frac1{a+j\omega}$ and ${\cal F}\{e^{at}u(-t)\}=\frac1{a-j\omega}. $
Adding,
$\displaystyle \frac1{a+j\omega}\frac{a-j\omega}{a-j\omega}+\frac1{a-j\omega}\frac{a+j\omega}{a+j\omega} =\frac{2a}{\omega^2+a^2}. $

${\bf H} (j\omega)=j\omega e^{j3\omega}$

Solution:
$j\omega\;{\bf X} (\omega)\Leftrightarrow\frac{dx}{dt}$ and ${\bf X} (\omega)\;e^{j3\omega}\Leftrightarrow x(t$+3).
Combining gives $\frac{dx}{dt}(t+3)$.

$h(t)=te^{-2t}\;u(t)$

Solution:
From entry #11 in Table 5-6, $\displaystyle {\cal F}\{te^{-2t}\;u(t)\}=\frac1{(j\omega+2)^2}. $
Inserting $\omega=2$ gives
$\displaystyle \frac1{(j2+2)^2}=\frac1{(2\sqrt{2}e^{j45^\circ})^2}=\frac1{8}e^{-j90^\circ }. $
Then $y(t)=\left(\frac{1}{8}\right) 8\cos(2t-90^\circ)=\sin(2t)$.

Solution:
${\bf H} (\omega)=e^{-j2\omega}$ which is the frequency response of a delay by 2.

Solution:
$\displaystyle {\rm Gain}=|{\bf H}(\omega)| =\frac{|j\omega-2|}{|j\omega+2|} =\frac{\sqrt{1+\omega^2}}{\sqrt{1+\omega^2}}=1. $
This is called an all-pass system.

Solution:
Consider each of the four candidate phase responses:

(a) LTI system $y(t)=x(t+3)$ has transfer function ${\bf H} (\omega)=|{\bf H}(\omega)|e^{j\phi(\omega)}=e^{j3\omega}$. Hence, the phase response is $\displaystyle \phi(\omega)=3\omega. $
Since phase response belongs to a viable LTI system, it is not the response to the posed question.

(b) The reversal property given by Eq. (5.130) requires that the transfer function satisfy $\displaystyle {\bf H}(-\omega)={\bf H}^*(\omega), $
which implies that $\phi(\omega)=-\phi(-\omega)$.
Clearly, $\phi(\omega)=3\omega^2$ does not satisfy the reversal property. Hence (b) is the answer to the posed question.

(c) LTI system $y(t)=x(t)\;*\;e^{-t}\;u(t)$ has transfer function $\displaystyle {\bf H} (\omega)=|{\bf H}(\omega)|e^{j\phi(\omega)}=\frac1{j\omega+1}\;. $
Hence, the phase response is $\displaystyle \phi(\omega)=0-\arctan\left(\frac{\omega}{1}\right) =-\arctan(\omega). $
Since the phase response belongs to a viable LTI system, it is not the response to the posed question

(d) LTI system $y(t)=\frac{dx}{dt}$ has transfer function $\displaystyle {\bf H} (\omega)=|{\bf H}(\omega)|e^{j\phi(\omega)}=j\omega. $
Hence, the phase response is $\displaystyle \phi(\omega)=\pi\;{\rm sign}(\omega). $
Since the phase response belongs to a viable LTI system, it is not the response to the posed question.

Solution:
By Parseval's theorem, $\displaystyle \frac1{2}\left(1^2+\frac1{3^2}+\frac1{5^2}+\cdots\right) $
is the average power of $x(t)$, which can be computed in the time domain as $\pi^2/16$ since
$\displaystyle |x(t)|^2=\frac{\pi^2}{16}. $
Then $\displaystyle 2\;\frac{\pi^2}{16}=\frac{\pi^2}{8}\;. $

Solution:
Eq. (5.28c) for computing coefficients $\{b_n\}$ of sine terms in a Fourier series is
$\displaystyle b_n=\frac{2}{T_0}\;\int_0^{T_0}x(t)\sin(n\omega_0t)\;dt, $
where here $T_0=2\pi$ is the period and $\displaystyle \omega_0=\frac{2\pi}{T_0}=1. $
So the desired integral is $\displaystyle b_3=\frac1{3}\;$, which agrees with direct computation.