Looks like the PMF is supposed to be
[tex]\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}[/tex]
which is kinda weird, but it's not entirely clear what you meant...
Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:
[tex]\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65[/tex]
Next,
[tex]\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25[/tex]
[tex]\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5[/tex]
[tex]\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2[/tex]
[tex]\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5[/tex]
[tex]\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}[/tex]
If [tex]Y=X^2+1[/tex], then [tex]X^2=Y-1\implies X=\sqrt{Y-1}[/tex], where we take the positive root because we know [tex]X[/tex] can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that [tex]Y[/tex] can take on the values [tex]1^2+1=2[/tex], [tex]2^2+1=5[/tex], and [tex]5^2+1=26[/tex]. At these values of [tex]Y[/tex], we would have the same probability as we did for the respective value of [tex]X[/tex]. That is,
[tex]\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}[/tex]
Part (5) is incomplete, so I'll stop here.
