The problem is as follows:
The diagram from below shows a pendulum. It is known that the magnitude of the tension of the wire is five times of that the weight of the sphere which is attached to. Acording to this information indicate True or False to the following statements:
I. The magnitude of the centripetal force which acts on the instant shown is $5mg$
II. The magnitude of the tangential force which acts on the instant shown is $0.6mg$
III. The magnitude of the total acceleration in the instant shown is $3\sqrt{2}g$.
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&FFF\\ 2.&FTT\\ 3.&FTF\\ 4.&TTF\\ 5.&TFT\\ \end{array}$
From my attempt I could spot is the centripetal force would be:
$F_{c}=\frac{mv^2}{R}$
The problem mentions:
$T=5mg$
From the sketch I made (and if my interpretation is correct)
Then:
$T=5mg=f_{c}$
Therefore the first statement is True.
But I'm stuck there. How can I find the other two?. Can somebody help me here?.
(I am answering under the assumption that the angle $37^\circ$ was given to you in the problem statement, which you didn't mention in the body of your question.)
Newton's law of motion for the bob can be written as $$ m\mathbf a_{net} = \mathbf F_{net} = \mathbf T + \mathbf F_{g}, \tag1$$ where $\mathbf T$ is the tension applied by the wire onto the bob and $\mathbf F_g$ is the gravitational force on the bob. We may choose our axes so that the origin is the pivot point, the $x$-axis is pointing downwards and the $y$-axis is pointing to the right, and adopt polar coordinates: this way, the position of the bob at all times is given by a pair $(r,\theta)$, where $r$ is the distance from the pivot, and $\theta$ is the angle with the vertical. (Observe that the wire has fixed length $\ell$, so the motion will be circular around the pivot, and the position of the bob is determined via the single variable $\theta$.)
Fix a specific value of $\theta$. Our choice of coordinates suggests that we decompose any vector acting on the bob into one radial component and one azimuthal (angular) component; in other words, we are choosing two special unit vectors acting whose foot is placed on the bob: one is $\mathbf e_r$, which lies on the line through the pivot and the bob, and points away from the bob, and the other is $\mathbf e_\theta$, which is perpendicular to $\mathbf e_r$ and thus tangent to the circular trajectory of the bob, pointing towards the direction of increasing $\theta$. (This is where this answer diverges from your attempt: you have chosen to decompose your vectors in a different way that is ultimately less helpful in solving the problem at hand.)
$\hskip2.5in$
All vectors acting on the bob can be decomposed in terms of these two unit vectors; in particular, we can decompose all forces appearing in $(1)$. $$\begin{split} \mathbf F_{net} &= \mathbf F_{net}^\perp + \mathbf F_{net}^\parallel = F_{net}^\perp \mathbf e_r + F_{net}^\parallel \mathbf e_\theta \\ \mathbf T &= \mathbf T^\perp + \mathbf T^\parallel = T^\perp \mathbf e_r + T^\parallel \mathbf e_\theta \\ \mathbf F_{g} &= \mathbf F_g^\perp + \mathbf F_g^\parallel = F_{g}^\perp \mathbf e_r + F_{g}^\parallel \mathbf e_\theta \end{split} $$
Let's figure out what the components of $\mathbf T$ and $\mathbf F_g$ are. A bit of trigonometry tells us that the radial component of the weight has magnitude $mg\cos\theta$ and is pointing directly away from the pivot, so $$\mathbf F_g^\perp = mg\cos\theta\ \mathbf e_r; $$ on the other hand, the azimuthal component of the weight has magnitude $mg \sin\theta$ and is directed toward the vertical ($x$-) axis, so $$\mathbf F_g^\parallel = -mg\sin\theta\ \mathbf e_\theta $$ For the components of $\mathbf T$, notice that the tension is always perpendicular to the circular trajectory, so $\mathbf T^\parallel = \mathbf 0$ and $\mathbf T^\perp = \mathbf T$. Since $\mathbf T$ is always pointing toward the pivot, i.e. in the negative radial direction, we have found that $$\mathbf T^\perp = \mathbf T = - \|\mathbf T \|\ \mathbf e_r . $$
Now let us rewrite the vector equation $(1)$ in terms of these components. Since $$F_{net}^\perp \mathbf e_r + F_{net}^\parallel \mathbf e_\theta = \mathbf F_{net} = \mathbf T + \mathbf F_g = -\|\mathbf T\|\ \mathbf e_r + mg\cos\theta\ \mathbf e_r - mg\sin\theta\ \mathbf e_\theta, $$ we may compare the coefficients of $\mathbf e_r$ and $\mathbf e_\theta$ on both sides to obtain the system $$ \begin{cases} F_{net}^\perp = -\|\mathbf T\| + mg\cos\theta, \\ F_{net}^\parallel = -mg\sin\theta \end{cases} $$ Notice that, due to our setup, the LHS of the first equation corresponds to the magnitude of the centripetal force $$\mathbf F_c = - \|\mathbf F_c\|\ \mathbf e_r$$ experienced by the bob; in other words, $$\|\mathbf F_c\| =\| \mathbf T\| - mg\cos\theta. \tag2 $$
In your case, you are being asked to analyze the situation where $\theta_\star = 37^\circ$ and $\|\mathbf T\| = 5mg$. For this particular choice of angle, trigonometry tells us that $$\cos 37^\circ = \frac 4 5, \qquad \sin 37^\circ =\frac 3 5,$$ so that equation $(2)$ becomes $$\|\mathbf F_{c}\| = 5mg - mg\cos\theta_\star = \frac{21} 5 mg, $$ which tells you that statement (I) is false.
Secondly, the magnitude of the net tangential force is given by the azimuthal equation to be $$\|\mathbf F_{net}^\parallel\| = mg\sin\theta_\star = \frac 3 5 mg, $$ which establishes (II) to be true. Finally, remembering that $$\|\mathbf a_{net}\| = \frac 1 m\|\mathbf F_{net}\| =\frac 1 m \sqrt{(F_{net}^\perp)^2 + (F_{net}^\parallel)^2 }, $$ we may calculate that $$\|\mathbf a_{net}\| = \frac 1 m \sqrt {\frac{450}{25}m^2g^2} = 3\sqrt2 g,$$ so (III) is also true.